Dynamics Checklists/ HInts

Process to solve problems

1.) Know the equation sheet like the back of your hand 2.) Identify Particle/Rigid body 3.) Identify the type of problem (Energy, Impulse/Momentum, FBD, Impact, Dependent Motion, Relative Motion, Absolute Motion ) 4.) Draw a map starting from given to unknowns 5.) Do not violate any principles 6.) Start with the coordinate frame, or possibly FBD & KD 7.) Use appropriate principle

Final Exam

1.) Know the equation sheet like the back of your hand 2.) Identify Particle/Rigid body Identify if a problem is a rigid body, or a particle. If its a particle, the first question you ask yourself is, Acceleration. If its a rigid body, the first question you ask yourself is, What kind of motion do I have? 3.) Identify the type of problem (Energy, Impulse/Momentum, FBD, Impact, Dependent Motion, Relative Motion, Absolute Motion ) What bubble (concept) would you put on this problem. For distance you would use WED, for time MIT, If they are connected than dependent motion. If its rolling then you have the kinematic equation. 4.) Draw a map starting from given to unknowns If it doesent seem right to you, look at it again. leave it go to another problem, then return and look at it again. The last thing that you want to do is violate any principles. Do not violate any principles. No partial credit, answer is completely incorrect 5.) Start with the coordinate frame, or possibly FBD & KD For the entirety of every concept in dynamics you need a FBD. For equations of motion you need the KD. You need to be very comfortable with FBD 7.) Use appropriate principle Hints*) -) Kinematics and kinetics kinematics is the study of position, velocity, acceleration. -) Particle -) You have the particle, it moves, it has some position, some velocity, some acceleration -) Whatever we did for the particle we also did for the rigid body. -) We have position, velocity, and acceleration for a particle. The first question you ask yourself is, what is the acceleration, Is the acceleration constant? or is the acceleration variable? -) Rigid Body -) Motion is the most important thing for the rigid body, what kind of motion did we have for this rigid body? -) Is this rigid body translating (rectilinear translation or curvilinear translation), -) is it rotating (Pure rotation about point O, Pure rotation about G), or is it General motion. -) General motion is = to relative motion -) Kinetics Kinetics is the study of forces and moments, your FBD, work energy, momentum-Impulse -) This class was divided into two sections, one was the particle and the other one was the rigid body. -) The only differene between the two, is that a particle has no dimensions. You can have a bus as a particle, as long as you neglect the rotation part of the bus. It becomes a particle. The entire mass is concentrated at one point -) If its a rigid body, you have dimensions, you have length, and you have radius -) Both particles and rigid bodies have mass -) Geomotry is only available in the rigid body. That's one way of distinguishing if you have a particle or a rigid body. -) For a rigid body, the mass center point 'G' is the most important Everything is based on that mass center point 'G' (Igomega , 1/2 Igomega^2) -) At the core of this class, is the coordinate frame. That's the first thing the professer will be looking for in the exam -) Their are three different types of coordinate frames. -) The rectangular coordinate frame, How do you do the position, the velocity, and the acceleration of the rectangular coordinate frame. -) The n-t coordinate frame, How do we know when to use the n-t? If you have a curved motion. The particle, or the rigid body is going through curved motion. and rho (rho), the radius of curviture is given to you Then more than likely you are going to use the n-t coordinate frame. -) If r and theta were given to you, then most likely you would use the r-theta coordinate frame. -) In any coordinate frame, regardless of coordinate frame we know how to determine the position? the velocity? and the acceleration of the particle. Using the equations you have in your equation sheet. -) You can also use the EOM (Equations of Motion), for each one of these coordinate frames. -) Coordinate frame always comes first. You first decide what your coordinate frame is. Then comes everything else. -) Particle Only mass, no geometry, no dimensions, no rotation Particle is position, velocity and acceleration -) The graphical method (Erratic) Where you have a graph of position vs time Or you can have velocity vs. distance. Or you can have acceleration vs. distance -) For a particle, the question of motion doesent show up because its only translation. We said no rotation for the particle -) For a particle the first question you ask yourself is, what is the acceleration? Is the acceleration constant, or is the acceleration variable. If the acceleration is variable, is the acceleration a function of time? Or is the acceleration a function of position. -) Learn which equations to use for constant acceleration, and what equations to use for variable acceleration. -) In variable acceleration, it can be a function of acceleration 'a', it can be a function of the distance 's', or the acceleration 'a' can be a function of time a(t). Is it a time based problem? Or a position based problem -) When you take the integral of both sides, then the equation on the left must be the same as the equation on the right -) If you're integrating over time, everything in their should be a function of time. If on the right side you are integrating over position, then everything in that expression should be a function of position. If acceleration, then everything should be a function of acceleration. -) That's the motion for a particle, In a particle its only translation. -) If the particle is moving just by itself. And it doesent come into contact with anything. Then distance -) If it's no contact, then the particle moves with a distance D, (WED) -) When can WED be simplified into COE? -) What is COE? -) If you have multiple particles, If the multiple particles are connected, then you have dependent motion. If the multiple particles are not connected to each other. if you have one plane moving, then another plane moving in a different direction. Their is relative motion going on between the two particles If your problem has distance, WED. If your problem has time MIT -) If their is time in your problem, MIT -) Projectile motion moition along the 'x', the acceleration along the 'x' is zero. The acceleration along the 'y' is plus or minus g depending on where your axis is, where your positive Y is -) Impulse-Momentum -) Conservation of Momentum. For momentum theirs conservation of linear momentum, and their is conservation of angular momentum -) If you have MIT, you have linear momentum Impulse and angular momentum impulse. a particle can have angular momentum -) For dependent motion -) When you have an impact problem, state 1 and state 2 are pre-defined for you -) For WED or COE, you can arbitrarily use any state, state 1 and state 2

Exam 2 Review

Check List*) 1. Coordinate frame (x-y), (n-t), (r-theta) 2. Free body diagram & Kinetic diagram 3. "WED", T1 + sumu1→2 = T2 4. "COE", T1 + V1 = T2 + V2. Datum? V1 = V1g + V1e ; V2 = V2g + V2e *V1g<or>, V1e = 1/2KS1^2, V2e = 1/2KS2^2 5. Impulse Moment DIagram "MIT" Linear Impulse Moment principle Angular Impulse Momentum principle "AMIT" 6. Units Professer will be looking for*) 1.) Coord frame ( x-y ) ( n-t ) ( r-theta ) If its an x-y problem, I will be looking for the x-y coordinate frame -)If you have a plane, it only makes sense to point the x-axis along the direction of the plane, and the y-axis 90 degrees to that. -)If you have a curved path, and if rho is given to you. It would make sense to use the n-t coordinate frame. -)If rdot, thetadot, rdouble dot, theta double dot, is given to you then it would make sense to use the r-theta coordinate frame. 2.) FBD & KD Will be looking for FBD and KD 3.) (WED) T1 + sumU1→2 = T2 When you use the WED equation, please make sure that you define the states. State 1 & state 2. State 1 is your initial state. State 2 is when the velocity is whatever. Spring force Us = -k/2 (s2^2 - s1^2) Where the S's are, the stretch of the spring 4.) (COE) T1 + V1 = T2 + V2 When you write down COE, Immediatly will look for Datum. Elastic energy = 1/2 k stretch 1^2 ; (V1e = 1/2 ks^2) Potential energy = mg * (distance from datum) 5.) MIT ( Linear Impulse momentum principle, Impulse-Momentum Diagram ) If you have time in your problem, think of MIT. If you have distance in your problem think of WED. AMIT ( Angular impulse momentum principle ) If you have rotation and time think of AMIT 6.) Units Hints*) -) If your problem asks for a force or an acceleration. The force, or the moment comes first. -) Once you have a force, that force results in an acceleration, velocity, or position -) If you have a moment, the moment results in angular motion, angular acceleration, angular velocity, and angular position. -) If your problem is asking for a force or acceleration? Think of using the FBD and the KD. -) The FBD and KD -) Remember, the free body diagram only forces. The kinetic diagram mass * acceleration (sumF = ma). -) If your coordinate frame is x-y, then you have (sumF = ma) along the X and (sumF = ma) along the Y. Remember you may not have an acceleration along the y -) The n-t coordinate frame, and the r-theta coordinate frame. -) Once you have the FBD and the KD, all you say is this is = to this vector by vector -) (sumF = ma) (x direction), sum of all forces along the x direction is equal to mass * acceleration along the x direction (sumF = ma) (n direction), sum of all forces along the n direction is equal to mass * acceleration along the n direction -) The convention for the n-t. 'n' always points torwards the center of curviture. and 't' is tangential -) Non-Impact distance -) no impact, no contact, no collision -) By now you should be able to look at a problem and know if its a contact problem? or a non-contact problem. -) If you have no impact, and you have distance. Remember WED. -) If you do use WED, make sure you define the states. state 1 & state 2 -) What is state 1 and what is state 2 -) What is the challenging think about WED -) Energy = 1/2m v1^2 + 1/2m v2^2. Then comes the work by all the forces in your FBD. -) The challenging part about WED is the work done by the spring force -) You have to do the stretch (s1), the stretch (s2), the work done by the spring is (-1/2 K s2^2 -s1^2) -) What is the stretch? The length in that state minius the free length. -) WED becomes COE, If you have no external forces and no friction. These are the two conditions. -) COE (Conservation of Energy) In COE make sure you put down state 1 and state 2. COE between the two states. -) When you use COE, the moment you write down the Potential Energy what do you have to worry about? The different types of Potential energies, Elastic and Grav. -) How many different types of potential energy do we have? Two, the Elastic potential energy and the gravitational potential energy. -) When you're writing down the gravitational PE, what should you define? The datum. -) The datum depends on wherever you want to put it, it's fixed to the ground. Below the datum the gravitational PE is negative. Above the datum, the gravitational PE is positive -) Can you write down the checklists, for each of these concepts? -) The moment you have an Impact, two or more particles coming into contact with each other? Then you can have linear Impact or rotational Impact. -) If you have an arm, that can rotate. You also have an external arm stretching out of this vertical arm. When this vertical arm moves, then the horizontal arm is also going to move. -) Rotation: Angular momentum, Angular Impulse -) Linear impulse and angular impulse. -) If you have an Impact problem? Remember to define the LOI (Line of Impact), which is 90 degrees to the common tangent. -) Along the line of Impact, you have COM and COR Along the y-axis you only have COM. -) If you have an Impact problem, you cannott use WED or COE. You have to use COR or COM. -) MIT ( Impulse-Momentum Diagram, Linear Impulse momentum principle) AMIT ( Angular impulse momentum principle )

18.5 Conservation of Energy

Check List*) 1. Distance problem, WED Look for a distance in your problem, if their is a distance think of WED. After you think of WED, ask yourself does COE apply? 2. Does COE apply -) No friction for non round objects Does it apply for non round objects, no friction -) Only rolling for round objects If you have a round object, then it must be rolling. The problem must state, or you must figure out if the round rigid object is rolling or sliding 3. Rolling object, Vg = Romega (Kinematic constraint) When you have a rolling object, remember the kinematic constraint. Vg (Velocity of mass center point g) is r multipleid by omega (Vg = Romega) 4. Only weight, spring, and normal forces in FBD In your FBD, you should only have weights, spring, and normal forces 5. No external forces or moments only then can you apply COE 6. Define coordinate frame and datum for Vg You need the datum for the gravitational PE. 7. Define states Define your states, your states can be anything that you choose. State 1 and State 2. 8. T1, V1, T2, V2 Kinetic energy in state 1, potential energy in state 1, Kinetic energy in state 2, potential energy in state 2 9. Solve for unkown (1 equation 1 unkown) COE Conclusion.) -) No external forces/ moments that do work -) No friction force/ rolling -) Tire rolling -) V1 = Vg1 + Ve1 -) V2 = Vg2 + Ve1 -) Vg → Datum -) Rolling kinematic constraint (Vg = Romega) Tricks*) 1.) Non round objects, one equation and one unknown Round object, two equations and two unknowns Of the two equations, one is going to be COE, and the other is going to be kinematic constraint. (Vg = Romega) Hints*) -) The work energy equation Kinetic energy in state 1 + sum of all the work done from state 1 to state 2 is equal to the kinetic energy in state 2 (T1 + sumU1-2 = T2) -) Simplification of WED, work energy simplifying work energy becomes COE. -) If theirs COE for a particle, then their has to be COE for the rigid body. -) Kinetic energy of a rigid body, rigid body general motion means translation and rotation. -) How many components of the KE is their? Translational KE and rotational KE. -) For rotational KE, its just 1/2 mass moment of inertia with respect to the point in rotation omega^2 (1/2(mv^2G + IG omega^2)) -) How do I get Io, if I have Ig? I use the parallel axis theorem. -) COE is only valid, no external forces, no external moments and no friction -) Rigid bodies can either be round, or non round -) If you have a round rigid body, and that rigid body is rolling what is the work done by the friction force if the rigid body is rolling? zero -) If you're rolling, than the work done by that friction force is zero. But if you are sliding, If you brake hard and the wheels lock. And you're sliding on the pavement. You are leaving rubber behind. youre leaving some of the tire behind, streak Thats where work is done -) If its a round object thats rolling, you can use COE as long as you dont have an external force -) your FBD will have only these forces. Weight, spring force, normal force, and rolling friction force. -) Only if youre rolling, the friction force due to rolling is fine to be in the FBD. Otherwise anything other than these four forces do work. Hence, COE is not valid -) The COE, is work energy that applies to the entire system -) If you apply it to one rigid body, of the entire system Then you have to take the FBD of that rigid body, seperate it from the entire system -) When you seperate the rigid body from the entire system, what do you have? reaction forces Then you have to determine the work done by those reaction forces -) When your using work energy for a rigid body, you are looking at an entire system -) You can apply work energy for a single rigid body Then you have to seperate that rigid body from that entire system, once you seperate it. What do you get? reaction forces. -) Conservation of energy says, the sum of kinetic energy of all the rigid bodies in your system If their is a rigid body, that moves translates or rotates It has energy. -) Sum of all the kinetic energy in state 1 + sum of all the potential energy in state 1 is = to sum of all the kinetic energy in state 2 + sum of all the potential energy in state 2 sumT1 + sumV1 = sumT2 + sumV2 -) The gravitational potential energy and the elastic potential energy -) The gravitational potential energy is just Mgh, h is the distance from the datum. -) When you simplify WED from COE. You must define the datum -) The potential energy is positive above the datum, and the PE is negative below the datum -) The elastic potential energy is 1/2KS^2. Since its 1/2K stretch squared. The elastic potential energy is always positive, the gravitational potential energy can be positive or negative depending upon your datum -) General motion is equal to pure rotation about the instantaneous center

12.9 Dependent Motion

Check List*) 1.) Count the number of ropes - Each rope gives one rope equation (L = constant) 2.) Define position coordinates from fixed reference/ datum lines, along the path of each particle. Different datum lines can be used for each particle, going through a different direction of motion If you have vertical motion, you have one datum If you have horizontal motion, you have a different datum 3.) Relate the position coordinates to the cord length. Segments of cord that do not change in length during the motion, may be left out. 4.) If a system contains more than one cord, relate the position of a point on one cord to a point on another cord. Separate equations are written for each cord. 5.) Differentiate the position coordinate equation(s) to relate velocities and accelerations. Keep track of signs! Tricks*) 1.) Define a reference for Horizontal Motion & a Different reference for Vertical Motion (Review rope pull down example), The point a goes through horizontal motion, and the point b goes through vertical motion. We're going to have two references. one for vertical motion and one for horizontal motion. (Review truck-furniture example), we're going to have two references. One for horizontal motion and one for vertical motion Remember horizontal motion reference, is a different reference to the vertical motion reference. reference is fixed to the ground, that's you're origin. The position of the particle is measured from the reference 2.) Reference is fixed to the ground You can fix your reference, to either of the fixed points. Even if they're not a particle, they just need to be fixed to the ground or the structure (thats fixed to the ground). (Example reference point's, Xa and Xb) (Position of 'b' to the horizontal reference, position of 'a' to the vertical reference). 3.) Position of particle measured from reference Position of particle a is measured from the reference. To the left(←) is positive(+) ; to the right(→) is negative(-) ; Towards the floor(↓) is positive(+) ; towards the ceiling(↑) is negative(-). 4.) Beware of direction of +-Vel -)If Va is positive(+) 5 m/s^2, (Va = 5m/s^2). Then the velocity Va is pointing in this direction ← If Va is negative(-) 5 m/s^2, (Va = -5m/s^2). Then the velocity Va is pointing in this direction → -) The velocity goes with the position of the particle measured from the reference. If Vb = +5 m/s, what is the direction of Vb? Since Xb is (+) going from the reference to the point. The Vb would be going down. If Vb = -5 m/s, what is the direction of Vb? The Vb would be going up Hints*) -) Dependent motion is the motion of multiple particles when they are some how connected. The key word is that they are connected by a 'tart', meaning (non slack) rope/ chain. Dependent motion: multiple particles connected by a tart (non-slack) rope, or chain -) In a rope-pull down machine, we have two objects. A bar (point A) is one particle, and Weight (point B) is another particle. They are connected by a tart (non slack cable) -) When a truck (point A) is pulling a piece of furniture (point B), over a pulley. You have two particles, The truck and the furniture. Both of these particles are moving in rectilinear motion. A & B is rectilinear (Individually). Individually they are going through rectilinear motion. However, their motion, position, velocity, and acceleration is not completely Independent of each other. They are dependent on each other. Meaning if you had the velocity of one, you could determine the velocity of the other. If I had acceleration of one, I could determine the acceleration of the other. If I had position of one, I could determine the position of the other. Position, velocity, and acceleration are dependent, meaning their's a relationship between both of them -) Another example of dependent motion is, a pulley. In our example we have two particles, particle b, and particle a. Both a and b (individually) are going through rectilinear motion. Motion along one axis. Motions A and B are not independent of each other because they are connected through a tart rope. -) Non-dependent motion, two particles connected via a slacked rope. a non tart rope. Here the motion of P and the motion of Q are independent as long as their is slack in the cable/rope. -) Remember when we were doing rectilinear motion? If you have an origin and a particle p. If you're particle is moving away from the reference (origin). The velocity Vp would be positive. If the particle is moving towards the reference, than the velocity would be negative. ("When moving on a straght road, If you're going foreward the speed is positive. But if you are going in the opposite direction, then the speed is negative") -) If you have two particles 'a' and 'b' that are connected via the rope? This is dependent motion, therefore a and b have dependent motion. Particle a goes through horizontal motion, particle b goes through vertical motion. -)I need two references, the reference for the horizontal motion, and the reference for the vertical motion. The reference has to be placed, so that its attached to the ground. (The pulley is attached to the ground), that is our reference point. For the vertical motion you can have the same point. But the reference is horizontal because you are measuring the vertical position. -) If Va = + 5 m/s, the direction it would move is to the left. If Va = - 5 m/s, the direction it would move is to the right. -) **Instead of writing Va = -5m/s ; write Va = 5m/s → ** -) Assumption is that the rope is tart (non-slack). -) When particles are interconnected by a cable, the motions of the particles are Always dependent -) If the motion of one particle is dependent on that of another particle, each coordinate axis system for the particles Should be directed along the path of motion

18.4 Principle of Work and Energy

Check List*) 1.) Distance → work-energy (WED) If its a distance related problem, remember WED 2.) sumT1 + sumU1→2 = sumT2 The sum total of all kinetic energy + the work done by all the forces in your FBD in state 1 & 2 is = to the sum total of all kinetic energy in state 2. 3.) Define state 1 & 2 Define your states, you define your states wherever you want. State 1 & state 2 4.) u1→2 : FBD (Ff does no work when rolling) Determine the work done by all of the forces in your FBD from state 1 & state 2. Remember friction force does not do any work, when you are rolling 5.) sumT1 & sumT2 You have the sum of all kinetic energy in state 1 and the sum of all kinetic energy in state 2 If you have pure translation, pure rotation or if you have translation and rotation or general motion. Depending upon the type of motion you have, your KE is going to be different Hints*) -) Last class we looked at the work of a force, and the work of a moment -) The work is used in the Work energy principle. What is the work energy principle? The sum of all the work done by all the forces in your FBD + the kinetic energy in state 1 is equal to the kinetic energy in state 2. sum T1 + sum U1→2 = sumT2 -) We have done sum of all the work done by all the forces. -) We're going to discuss the 2nd part of work energy, which is kinetic energy. -) The kinetic energy of a rigid body is 1/2mvg^2 -) What is the kinetic energy of a particle? 1/2 mv^2 if its 1/2 mv^2, what would be the kinetic energy for a rigid body? -) Why the mass center point G, and not some other point? The mass center point G represents the entire rigid body. -) What is the main difference between a particle and a rigid body? A particle doesent rotate, it only translates. A rigid body is going to translate and rotate, it can have an omega -) Back when we were doing particle dynamics, we never talked about rotational speed, we never talked about rotational speed of a particle. -) Does the particle have to be a point particle, like an atom or an electron? no, it can be a huge bus, truck, plane. That can be a particle too. -) As long as you concentrate that entire mass, at that mass center point 'g'. that mass centerpoint g, when completed. Doesent do any rotation. -) When you look at a car, and a car is moving striaght. It still has some kind of rotation. If you can ignore all of those rotations, The car is a particle -) You can look at it as a particle, if you ignore rotation. -) The moment you have rotation, now you have rigid body -) The rigid body has both translation and rotation. you are going to have a kinetic energy from translation (1/2mvg^2). If the translational KE, is 1/2mvg^2. Mass* the ^2 of linear velocity. -) What do you think would be the rotational KE of a rigid body? When you have mass for the particle that corresponds to what for rotation? Mass moment of Inertia Mass on a particle, corresponds to Mass moment of inertia (Ig) or a rigid body. -) Linear velocity v corresponds to omega. The rotational KE is 1/2 IG omega^2. ( 1/2mvg^2 + 1/2IGomega^2 ), you have two states. State 1 and state 2 -) If you look at the mass center point, it may have a velcoity vg1, and an omega1 in state 1. In state 2 it has a different Vg2, and omega2 That is the work energy for a rigid body. -) Particle no rotation, so omega 1/2mv^2 Rigid body has both tranlation and rotation, so you have 1/2mvg^2 + igomega^2 -) When you are doing the KE for a rigid body. remember the KE in a rigid body has two parts. Translation and rotation. Translation is 1/2mvg^2 ; Rotation is 1/2Igomega^2 ; (M becomes Ig (Mass mom. of Inetia) ; Vg becomes omega) -) Remember for a rigid body, the most important point on a rigid body is the mass centerpoint G. That is the representative point for the entire rigid body. -) In order to compute the KE, you must be able to tell if this rigid body is going through translation, curvilinear translation, rotation. -) How many types of rotation do we have? Two, rotation about the mass centerpoint G, and rotation about a different point 'O', or 'P', or 'K'. Pure rotation -) General motion is the addition of translation & rotation -) If you have a rigid body going through pure translation, How many types of translation do we have? Rectilinear translation and translation along a curve (Curvilinear) -) For translation it is 1/2mvg^2, for rotation you have 1/2Ioomega^2 If it's rotating about the point 'O'. if its rotating abouut a point 'G' then 1/2Igomega^2 -) General motion Is general motion pure rotation? No, General motion is pure rotation if you use the Instantaneous center of rotation 'IC' -) Instanatenous center or rotation was a method to convert general motion into pure rotation. -) If you look at general motion as pure rotation about the IC, the kinetic energy is the same as rotation 1/2 Youre rotating about IC therefore 1/2 Mass moment of inertia with respect to the IC * omega^2 -) If you look at general motion as translation + rotation. Then that's 1/2 mvc^2 + 1/2 Igomega^2 -) You have the sum of all kinetic energy in state 1 and the sum of all kinetic energy in state 2 If you have pure translation, pure rotation or if you have translation and rotation or general motion. Depending upon the type of motion you have, your KE is going to be different -) Kinetic Energy : T1 + sum U1→2 = T2 -) The work energy principle is a total system. A total system, not individual links

19.1 - 19.2 Principle of Linear and Angular Momentum

Procedure*) 1.) Time in problem? If yes, MIT If you have time in your problem, think of using MIT 2.) Define coordinate frame + FBD Define your coordinate frame, then your FBD 3.) What kind of motion? Ask yourself what kind of motion does the rigid body go through Translation Rotation General Motion 4.) Translation If its translation, then rectilinear and curvilinear translation Linear-momentum Impulse principle 5.) Rotation Pure rotation Angular momentum Impulse principle 6.) General Motion Linear momentum and the angular momentum impulse principle Linear-momentum Impulse principle and Angular momentum Impulse principle 7.) Max 3 equations At a max you will have three equations and three unknowns solve for unknowns (Max unknowns = 3) Kinematic constraints (Vg = omegaR) Tricks*) 1.) Distance, WED (Work-Energy-Distance) If you have a distance related problem, think of WED 2.) Time, MIT (Momentum Impulse Time) If you have a time related problem, think of MIT 3.) Rotation + TIme, Angular momentum principle If you have a rotation problem, If you have rotation and time Think about using the angular momentum principle 4.) Total of three equation for planer motion. Two equations for Linear Impulse Momentum, One equation for Angular Impulse momentum 5.) If the problem asks for velocity, use the linear momentum principle Linear momentum in state 1 + sum of all the forces integrated over time from t1 to t2 is = to the linear momentum in state 2 Hints*) -) Distance is WED -) If you have time in your problem (MIT) Momentum Impulse Time -) Linear Momentum and Angular Momentum -) What is momentum for a particle? Mass * Velocity (mv) -) What is the kinetic energy of that particle? -) What is the linear momentum of a particle? Mass * velocity -) The angular momentum is about a certain point, kind of like the moment is about a point. -) The angular momentum is the moment of the linear momentum (mv) -) What is the moment of a force? r cross f, (r x f) -) The angular momentum is r cross the linear momentum (r x mv) -) If I were to bring this from the particle world, to the rigid body world Then the linear momentum (mv) becomes mass * the velocity of the mass centerpoint G. -) The linear momentum of a rigid boyd is = to mass * velocity of the mass center point G. -) The angular momentum of a particle is (r x mv) What would be the angular momentum for a rigid body? r cross m Vg. (r x mvg) Because 'g' is now the important point on a rigid body. -) On a particle everything is concentrated on a single point -) In a rigid body you have an infinite number of points. Their are two equations for the angular momentum, I * omega (Iomega) Linear momentum is mass * velocity (mv) -) Angular momentum is the analog of mass in the angular world is the moment of inertia. The analog of velocity in the linear world is omega (Iomega) -) If you want the angular momentum about Remember for angular momentum you have to specify the point. Exactly like the moment -) For the moment you have to specify the point about which you wanna take the moments of. -) Remember, the first thing you have to answer to is. What kind of motion do you have -) Their are three types of motions, Pure translation If you have pure translation, then the linear momentum is the mass * the velocity of the mass center point 'g' (L = mvg) -) Pure translation, no omega. Iomega is zero -) For pure rotation, you have to differentiate if you have rotation about the mass centerpoint g, or do you have the rotation of a different point o -) If you have rotation, about a different point 'o' The linear momentum, is mg. The angular momentum is the axis of rotation Ioomega -) Io, using the parallel axis theorem. from Ig plus mass * v^2 V^2 is the distance from g to the center of rotation -) If you have pure rotation, about the mass center point g. Then g doesent move, if you're rotating about the mass centerpoint g. Then g doesent move -) If g doesent move, then the velocity of g is zero hence, the linear momentum of the rigid body under pure rotation about the mass centerpoint g is zero. -) If you have omega, because you have rotation. Hg = Igomega -) What you have on the left, must match what you have on the right -) If you have o on the left, you have to have o on the right -) General motion In general motion, the type of motion that you have is translation and rotation. -) For translation and rotation, the linear momentum is mass * the velocity of the mass centerpoint g. -) General motion is also = to rotation about the instantaneos center -) If youre rotating about the instantaneous center, then all you have to do is -) If you want to determine the angular moment about the instantaneous center of rotation. That's I Ic omega. (MIc = Icomega) -) If you want to determine the angular momentum about g, then its Igomega. -) (Iic = Ig + md^2), Iic is parallel axis theorem = to ig + mass multiplied by the distance squared The distance of the mass center point g to the instantaneous center. -) If you have translation and time, then use the linear momentum principle. The linear momentum is state one + the sum of all the forces integrated over time from t1 to t, is = to the linear momentum in state 2 -) In these type of problems. You have force, velocity, and you have mass -) The angular momentum principle The angular momentum in state 1 about point o + the sum of all the moments about the same point o over time from t1 to t2 is equal to the angular momentum of that same rigid body about o is state 2. -) Where would you use Momentum impulse principle. First, you would use it if you have time in your problem. Second, you would use it if you had any kind of contact. If you have a force contacting another rigid body, that where you would use momentum impulse If you have two rigid bodies coming into contact, then momentum impulse -) For momentum impulse, the states are pre defined for you. state 1 is right before contact, state 2 is right after contact. That contact is happening, for only a short duration of time -) This is the same thing we did for particles, we're just expanding it for rigid bodies. -) When you expand the equations for particles, to a rigid body. The most important thing is the mass center point g -) Linear momentum is a vector, mass * velocity velocity is a vector. Then linear is a vector and so is force. -) Anytime you have a round object, and if that round object is rolling on the ground. The instantaneous center of rotation is mass center point . Vg = romega -) Linear momentum is the mass centerpoint g -) The centerpoint of the rigid body doesent have to be mass center point Linear momentum is m * velocity of the mass center point g -) The angular momentum is I about the point * by omega About which point? the point about which you are taking the sum of all the moments.

12.7 (n-t Coordinates)

Tricks*) 1.) Speed increasing by 4 m/s^2 => at = 4 m/s^2 If the problem says the speed is increasing by 4 m/s^2, that means the tangential acceleration is 4 m/s^2. -)Speed decreasing by 4 m/s^2 => at = -4 m/s^2 If the problem is decreasing by 4 m/s^2 then that means the tangential acceleration is -4 m/s^2 -) If you see the phrase "The speed is increasing by", they're talking about the tangential acceleration 2.) Speed constant => at = 0 If you see the phrase "The speed is constant", The tangential acceleration is zero. 3.) Radius or y(x) + curved path → n-t If the problem gives you radius, or y as a function of x + the curved path. Use the n-t coordinate Hints*) -) Coordinate frame is the most important thing for any dynamics problem -) Forces and moments are all with respect to the coordinate frame. -) Make a habit of defining the coordinate frame -) Their are three different types of coordinate frames that we're going to be looking at in this class. x-y coordinate frame, n-t coordinate frame, and Polar coordinates. These different coordinate frame's are used to simplify a dynamics problem. Sometimes it's easier to use the x-y, or the n-t, or the r-theta (Is the particle moving along a straight line? x-y? or a curve?) -) When do we use the x-y? When you have Rectilinear, Rectangular, and Projectile motion along two axis. -) You want to use n-t coordinates, whenever your particle is moving along a curve. If you have a curved problem use n-t. -) n-t coordinate frame is a coordinate system that is located on the particle and it moves with the particle -) In x-y coordinate system the origin is fixed. -) In polar coordinate system the origin is fixed -) The n-t coordinate system moves with the particle. In n-t coordinate system, 't' stands for tangential, 'n' stands for normal. -) t is always going to be tangential to the path. n is always going to be 90 degrees to that tangent -) The n axis always points towards the center of curviture of that particular curve at that particular point. -) For curved path problems we are going to be using the (n, t, b coordinates) -) The velocity vector is tangential to the path. -) The acceleration is going to have two components -) One component of the acceleration vector is tangential, the other component of the acceleration is normal to the tangential. Their are two components of the accel. -) The velocity vector, and the tangential acceleration are in the same direction along the tangent (Look at n,t,b coord) -) For normal acceleration all you have to do is take the speed sqrt(x^2) + y^2, speed^2/ radius of curviture -) If the particle is moving on a circle, the radius of curviture is the radius -) If the particle is moving on a curve, the radius of curviture is defined by (n,t,b coordinates) Equation rho equals. This is how you find the tangential acceleration, then use the speed to determine the normal acceleration. -) ("If you're driving on a curve, and you're speeding. That means the tangential acceleration is positive") -) Tangential acceleration is the change of your speed with time. -) Normal acceleration is your speed squared and divide it by the radius of curvature for that particular point on the curve. That becomes your normal acceleration. -) When you're driving on a straight line, your speed is increasing 50,51,52 mph. What is your acceleration? Its the tangential acceleration. When you're driving on a straight line the tangential acceleration is the same as youre acceleration. -) How many components of accleration do I have when im driving on a straight line. The radius of curviture on a straight line is Infinity. If someone draws a circle from infinity, that line is going to be straight. -) V^2 divided by infinity, the normal acceleration for rectilinear motion is zero -) The tangential acceleration can be constant or it can be varying with time -) If you have variable accleration, two components, two accelerations, tangential, normal, velocity only one component along the tangent -) The units for angle is radians, not degrees -) Normal acceleration points towards the center of curviture -) Absolute value of the normal acceleration is speed^2/ radius of curviture. (v^2/s) -) The tangential component of the acceleration is tangent to the curve (at = v dot) or (at ds = vdv) -) The normal component or the centripetal component, the normal acceleration or the centripetal acceleration. an = v^2/p (normal acceleration = speed^2 divided by radius of curvature) -) The magnitude of the acceleration, the total acceleration is sqrt(an^2 + at^2 ). (Acceleration is the square root of normal acceleration^2 + tangential acceleration^2) -) When you're moving along a straight line, the normal acceleration is zero. The tangential acceleration is the change in speed -) ("Next time your driving down the road, if its a straight line. You're tangential acceleration is something and your normal acceleration is zero. Or, if your in a curve, then tangential acceleration is the change of speed in your speedometer. Your normal acceleration is the same speed from your speedometer v^2/p ") -) Normal acceleration, when you're driving on a curve. You find that your body is being pushed outward. That's the normal acceleration doing that. That's the negative of the normal acceleration -) Normal acceleration points toward the center of curviture, and you're being pushed away from the center of curviture -) That's the centrifigal force, the negative normal acceleration multiplied by your mass -) an = v^2 / p (p is radius of curviture) If you have a nice perfect circle, the radius of curviture is the radius of that circle -) If you have a curve y=f(x), (y as a function of x) Then the radius of curviture is given to you by rho = [1 + (dy/dx)^2]^3/2 / |d^2y dx| -) When your speed is increasing, your tangential acceleration is greater than zero. -) When your speed is decreasing, your tangential acceleration is less than zero Even when driving down straight road, your tangential acceleration is either greater than zero or less than zero. The moment you go on a curve, then you also have a normal acceleration. -) Velocity always tangential to the path -) Acceleration two components tangential to the path and normal to the path -) The normal acceleration always points toward the center of curvature -) If a particle moves along a curve with a constant speed, then its tangential component of acceleration is Zero -)The Normal component of Acceleration represents The Time rate of change in the Direction of the Velocity

19.3 Conservation of Momentum

Check List*) 1. Do you have collision or Impact? If yes go to step 2 2. You are analyzing what happens during collision/ Impact 3. What type of motion -) If motion translation and sumFi = 0, then use conservation of linear momentum (sumLi = sumL2) -) If motion rotation and sumMo = 0 , then use conservation of angular momentum (sumHo1 = sumHo2) 4. If sumFi does not equal to zero, and time is in the problem, use Principle of Linear Momentum 5.If sumMi does not equal to zero, and time is in the problem, use Principle of Angular Momentum Tricks*) 1.) Look for axis of rotation If its a rotation problem, look for the axis of rotation. Once you know where the axis of rotation is ask yourself during impact, if the sum of all the moments about that axis of rotation is zero. If it is zero then you have conservation of angular momentum. If its not zero, and you have some force that's causing a moment about the axis of rotation. Than that force is going to accelerate, that rigid body. Conservation of angular momentum is out. The angular momentum of the rigid body with respect to the point of rotation in state 1, is equal to the angular momentum in the same rigid body with the same point. 2.) Is sumM Axis of rotation = 0? If yes COM Hints*) -) Momentum Impulse (MIT) -) If we have translation, If you only have external forces in your system then you have translation motion along the x, and motion along the y -) If you have an external moment, then that rigid body is going to rotate. -) Translation in time, gives you the linear momentum principle The impulse of a force, is the integral of that force over time The impulse of a moment, is the integral of that moment over time. -) The linear moment of a rigid body, mass multiplied by the velocity of the mass centerpoint of that rigid body mv is momentum. In the angular momentum. m becomes I, and v becomes omega. -) You only have one axis of rotation. -) For angular momentum, just like the moment you need a reference point -) You only have axis of rotation, in and out of the plane -) The most important part, in all of these principles. Is to determine what type of motion you have. If you have tranlation, then you have some kinetic energy If you have rotation, then you have some different kinetic energy If you have general motion, you have a different kinetic energy. -) If you have translation, and if the sum of all the forces is equal to zero Then you have conservation of linear momentum -) If you have rotation, and the sum of all the moments about the axis of rotation is zero Then you have conservation of angular momentum -) An example of conservation of linear momentum is If you have any kind of impact problem, any kind of a collision problem. Any kind of a contact problem Think of conservation of linear momentum -) If you have translation, and they are coming into contact with each other. Conservation of linear momentum. -) Conservation of linear momentum, only if you consider both particles as a system -) Conservation of linear momentum, shows up wherever you have impact -) When you kick a ball, the ball is at rest and you kick that ball. Their is contact happening, its a collision between your leg and that ball -) If their is contact happening, if you look at the entire system. The ball and the foot as one system. Then the force that the foot puts onto the ball is equal and opposite to the force the ball puts onto the foot. -) If you look at it, as one system. Then those reaction forces cancel out. You dont have any external forces acting you have mg, except mg is a non impulse force -) Therefore you have conservation of linear momentum -) If youre hitting a golf ball, with a gold club Contact, their is impact going on. Impact is conservation of linear momentum If and only if, you look at the golf club and golf ball as one system. -) The moment you seperate it, If you seperate it and you look at the FBD of the golf ball you have the weight of the golf ball, you have the normal force of the golf ball, and you have an external force that the golf club puts on the golf ball -) The moment you seperate these two objects that are colliding with each other. COM goes away, then comes the principle of linear momentum and impulse. -) If you look at it together as one, the forces cancel each other out. Sum of all the forces is zero. By definition, you have conservation of linear momentum. -) Only for a short duration of time, is your foot in contact with the ball. We're only looking at that short duration of time, sum of all the forces is = to zero. For that short duration of time, you have conservation of momentum.

14.1 - 14.3 Work-Energy Principle

Check List*) 1. Does the problem contain Force, Distance, and Velocity 2. If the problem contains Force, Distance, and Velocity then use principle of work-energy 3. Define the two states 4. FBD (KD not needed) 5. Compute work done by all forces in FBD 6. Compute the T1 & T2 of all particles in system 7. Use work-energy principle Tricks*) 1.) Does your problem have Force, Distance, and Velocity? WED 2.) Define two states 3.) Look for Lo For the work done by the spring. The first thing you have to do, Is look for Lo which is the unstretched length of the spring. What is the unstretched length of a spring? When you put a spring on a table, and apply no forces to it. whatever length that spring is, that's the unstretched length of the spring 4.) (S1 = L1 - Lo) and (S2 = L2 - Lo) If i take a spring, and I compress it to some length. Then the stretch is the the length in state one minus the free length. (S1 = L1 - Lo). The stretch in state two, the length is state two minus the free length. (S2 = Lo - L2) 5.) Us = -1/2 k (s2^2 - s1^2) (Work of the spring = - 1/2K stretch in state two^2 minus stretch in state one^2) Hints*) -) If you have two states, you have to define two states -) the first thing you wanna do is define two states -) before doing that, does your problem have Force, Distance, and Velocity? -) If your problem has Force, Distance, and Velocity. Think of WED (Work Energy Distance) -) If you have distance, work energy is the principle you should be looking at -) once u have defined the WED, the second step is defining your two states. state one and state two. -) The mother of all equations in dyanmics is, F=MA (sumF→ = ma→). That is the mother of all equations in dynamics -) If you have some sort of path, you have a particle moving on this path. This particle is subjected to a force F. You can see the force f has a component that is tangential to the path, and has a component that is normal to the path. -) Which component of that force does work? only the component that is in line with the motion Only Ft, this is going to be Fn, this is going to be Ft. Ft is along the tangent. Only Ft does work -) Sum of all forces equals mass times the acceleration of the tangential direction. sumFt→ = mat→ -) What is the tangential acceleration, (mat) multiplied by atds = vdv? at (tangential acceleration) is vdv/ds. sumFt = mat = m(vdv/ds) If I bring the ds to the other side, What do I have? M and M will cancel out. I have ds/at = vd. atds = vdv => sumFtds = vdv -) Now we have the mass as well, I can take the integral of both sides. what do I have? -) Integral (s1 to s2) sumFtds = Integral (v1 to v2) mv dv = m/2 [V2^2 - V1^2] Integral (s1 to s2) sumFtds is the work from state one to state two (u1→2) V2^2 is the KE in state 2, V1^2 is the KE in state 1. -) What does the principle of work & energy say? The work done from state one to state two plus the kinetic energy in state 1 is equal to the kinetic energy in state two -) For this problem, We need the FBD, and the work done by ever force in your FBD -) Work is the component of force, along the direction of motion, which is Fcos theta ds (Uf = Integral Fcos theta ds), -) if its constant force then (Uf = Fc costheta) deltas ) (The constant force comes out of the integral, costheta multipleid by deltas). The work done by the weight (Uw = -WdeltaY), The work done by spring (Us = -(1/2 ks2^2 - 1/2 ks1^2 )), The work done by couple moment (Um = M delta theta) -) What we've seen so far is from the mother of all equations, (sumF→ = ma→), we have the principle of work & energy. Principle of work and Energy ( T1 + sumU1-2 = T2 ) ; ( sumT1 + sumU1-2 = sumT2 ) -) If you have a force acting on a particle, then only that component of the force, which is Fcos theta. If this is theta, Fcos theta does work. Fsin theta does no work because its 90 degrees to the motion. -) Work done by gravity -) Remember gravity is always pointing straight down. mg is a force, so if the particle moves from state one, to a different height (h) state two -) Then the work done by the weight vector, is it positive or negative. If its going straight down from state one to state two? Work is force multiplied by distance. Is the force and the distance in the same direction? yes If its going straight down from state one to state two, It is positive. (u) is the symbol for work. Ug is the work done by gravity. -) Ug = + mgs (Work is equal to force times distance traveled) This is going from state 1 to state 2 -) If I were to take the particle from state 2 to state 1, from bottom to the top. It's going to be against gravity so the work is negative. (Ug = -mgs) -) What does it mean when the work is positive? or the work is negative? when you're going from state 1 to state 2, the particle is giving you work. -) The potential energy, It means that particle has some energy when it's at some height away from the floor. When you drop that energy, it drops to the ground. That energy is what causes the particle to bounce off the floor, its giving you work. -) When Ug is negative, that means you have to put work into the particle to raise it from state 2 to state 1. -) Negative(-) work, means you put work in. Positive (+) work, it gives you the work. -) Spring work -) The first thing you have to do, for the work done by the spring. Is look for Lo which is the unstretched length of the spring. What is the unstretched length of a spring? When you put a spring on a table, and apply no forces to it. whatever length that spring is, that's the unstretched length of the spring -) If i take a spring, and I compress it to some length. Then the stretch is the length in state one minus the free length. (S1 = L1 - L0) -) The work done by the spring, the definition is - 1/2K stretch in state two^2 minus stretch in state one^2. (S2^2 - S1^2) Us = -1/2K (S2^2 - S1^2) -) The work done on a spring by a particle attached to the spring. (Up = 1/2K (S2^2 - S1^2)) -) The S in spring stands for stretch, not length Stretch is defined as, the length in that particular state minus the free length. (L1 - L0) -) If I take the spring, and I expand it, then its (L2 - L0) length minus the free length. -) If I have a spring, dont do anything with it. That's the free length. If I expand it, then that's the total length of that spring in the expanded state minus the free length. meaning I have stretched it by this much. -) If I take the spring, the length of the spring. and I compress it, then it's L1 minus the free length of the spring. -) Remember the stretch is the length of that spring in that particular state minus the free length. -) Regardless if you're stretching it, or compressing it. The definition of the stretch, is the length of the spring compressed to a stretch. Minus the free length. That's the definition of stretch -) You're either stretching at positive, or you're stretching at negative -) Then apply this equation. (Us = Work of the spring) ; (Us = -1/2 k (s2^2 - s1^2)) -) Apply these three step. Look for Lo ; (S1 = Lo - L1) and (S2 = L2 - Lo) ; and Us = -1/2 k (s2^2 - s1^2) -) Spring unstretched -) Lo, unstretched in state one is zero -) Stretch is the length - the free length. -) In state 1. the length is equal to the free length. (S1 = L1-L0) In state 2, you are compressing the spring. The spring length has gotten shorter. (S2 = L2 - L0) -) What is the work done by the force F? Zero -) If a particle is moved from 1 to 2, the work done on the particle by the force, FR will be Integral S1 S2 Sum of F1 ds

13.6 Equations of Motion: (Cylindrical) (r-theta Coordinates)

Check List*) 1.) Curved path ? Is r, theta, rdot, thetadot given in the problem ? Is the path that you're talking about curved? If it is curved, do I have r, theta, rdot, thetadot? If I do then most likely the r-theta coordinate frame is what you should use. 2.) FBD and KD FBD and KD don't change, unless you use the r-theta coordinate frame. Their is a rule that you have to apply, the normal force is always 90 degrees to the tangent of the motion. Look at the motion, its basically tangent to the velocity. The normal is perpendicular to the velocity. 3.) Pay attention to convention for r-theta coordinate system 4.) Is radial direction 90o to tangent of motion ? If not, use psi If the radial direction is not 90 degrees to the tangent of the motion. their is an angle you must use. Decompose the friction force and the normal force into the r component and the theta component. psi comes into the picture if the radial direction is not 90 degrees to the tangent of the curve. 5.) psi exists if radial direction is not 90 degrees to the velocity→. 6.) EOM (Equation of Motion) Convention: Using r-theta coordinate system*) 1.) N is Always perpendicular to TANGENT of CURVE The normal force is always perpendicular to the tangent of the curve. If you have a particle thats moving on this curve. on the curve you have r and theta. Let's assume you have a bead with a hole in it, and its moving on that arc. The normal force direction, all you have to do is take the tangent, the tangent of the curve at that point. The normal force is always going to be 90 degrees to that tangent. In this particular case, the normal force is going to be 90 degrees to the tangent. The friction force is always going to be along the tangent, opposing the motion. Friction is going to oppose the motion. Friction force, what direction. The friction force is in the direction of the tangent. The friction force and normal force will always be 90 degrees of each other. They are always going to be perpendicular. 2.) v is always in THE TANGENT Direction The velocity is always tangent to the path, the velocity will be in the direction along the tanget. 3.) Friction FORCE Always ALONG TANGENT opposing velocity Friction force is along the tangent, opposing the velocity. Hints*) -)The last coordinate system polar coordinates, we did the x-y, we did the n-t, and we did the polar coordinates -) Remember force or acceleration is FBD or KD. -) In the sum of all forces = mass multiplied by acceleration (sumF→ = ma→), you take all the forces in your FBD, you decompose them in the appropriate directions equal to mass multiplied by the acceleration of that direction. -) (r, theta, z coordinates) look at the radial acceleration (ar), theta acceleration (atheta), and acceleration of the z direction (az). -) The same thing we did for the n-t coordinate frame. sum of all the forces in the radial direction (thetaFr). What is the radial direction for polar coordinates? So If i have a horizontal and vertical axis. The radial direction of the point p, It would be radially out from the origin, direction of the radial direction (ur). 90 degrees to the radial direction is the theta direction (utheta). -) When would you use the R-theta coordinate frame? when you have a curve, and If r, r dot, theta, theta dot is given to you. Then most likely, It makes sense for you to use the r-theta coordinate frame. -) Problem where you use radial coordinates theta dot is givem, most likely ill be using the r- theta coordinate. Ill place a particle on the rotating disk. That particle is going to move along some curve. tangent of the curve 90 degrees to that tangent is this angle. -) Remember, when we do the sum of all forces is equal to mass times acceleration (sumF→ = ma→), we have to take every force in our FBD, and decompose it into the r direction and the theta direction. -) Just like you had FBD in the n-t, you took all the forces in your FBD and decomposed it into the n direction, and the tangential direction. -)How to decompose Ff & Nc along r & theta direction? Lets assume that you have a particle that's moving on this arc, r is given, theta is given, theta dot is given so Im going to use the r-theta coordinate frame. -) Remember the normal force is 90 degrees to the velocity vector The velocity vector is along the tangent, 90 degrees to the velocity vector or 90 degrees to the tanget. Same thing Is the normal force -) Friction force is along the tangent, opposing the velocity. You might also have an external force. -) This is your FBD, how would you decompose all the forces of your FBD into the r and the theta component. -) If you want to decompose n in the r direction and the theta direction. Do you agree that you would need this angle? without that angle, you cannot decompose N into its -) This is the radial direction, what is the theta direction. If r is that way, what is the theta direction? 90 degrees to that Its the x-y, n-t. x is 90 degrees to y. -) If r is given to you, then you know what the direction of theta is. -) The direction of theta is 90 degrees counterclockwise. thats the direction of theta -) N has a component along the theta direction, and the radial direction. Is the radial component of n, positive or negative? Its negative because nr is in the opposite direction of the radial direction. -) Is n theta positive or negative? positive. what about the friction force, the friction force can also be decomposed into a component along the radial direction. Thats F radial, and the other one is F theta. Is the radial component of the friction force positive or negative? negative Is F theta positive or negative? Its negative because the positive theta is in that direction, therefore its negative. Do you see how the forces are decomposed? Do you see when the component of a force is positive? and when a component of the force is negative? The radial direction is from the origin to the point of interest. Then 90 degrees to that, counterclockwise is the theta direction. Just like we have x and y. Their is this angle psi, it is defined as tanpsi = rdtheta/dr = r/(dr/dtheta) -) psi is an angle, that was defined so that you could break up the normal force and the friction force in its radial component and its theta component. IF the radial direction is not 90 degrees to the direction of the velocity. then you have a psi in the equation. -) The radial direction and the tangent are not 90 degrees to each other. just because of where the particle is at the particlular time. The radial direction makes an angle, a non 90 degree angle with the tangent. The moment that happens you must define a psi. -) psi is the angle between the radial direction and the tangent P.s*) -) If i know what psi is, then what is that angle? (90-psi). Thats why psi is needed. -) The normal force which the path exerts on a Particle is always perpendicular to the Tangent to the path -) When the forces acting on a particle are resolved into cylindrical components, friction forces always act in the __________ direction? Tangential

17.2 - 17.3 Equations of Motion (Translation)

Check List*) 1.) Establish an (x-y) or (n-t) inertial coordinate system and specify the sense and direction of acceleration of the mass center, aG. 2.) Draw a FBD and kinetic diagram showing all external forces, couples and the inertia forces and couples. 3.) Identify the unknowns. 4.) Apply the three equations of motion (one set or the other): sumFx= m(aG)x ; sumFy = m(aG)y ; sumFn = m(aG)n ; sumFt = m(aG)t ; sumMG = 0 ; or sum MP = sum (Mk)P 5.) Remember, friction forces always act on the body opposing the motion of the body. Tricks*) 1.) Pure translation, α = 0 ; Pure translation (rectangular) α rotational acceleration is zero. Its moving along straight line Some rigid body moving along a straight line 2.) If problem is asking for force or acceleration, use FBD or KD If the problem is askiing for force, or acceleration, think of FBD or KD. 3.) These problems can only have 3 unknowns, because you only have 3 equations. Sum of all the forces in the x, sum of all the forces in the y, and sum of all the moments. You only have one moment equation because you have rotation about one axis coming out of the paper. 3 equations, 3 unknowns. Hints*) -) Mass moment of inertia is valid when you have rotation General motion has rotation, and pure rotation has rotation. Thats where we would use the mass moment of inertia. -) Their are two types of Translation, rectilinear and curvilinear -) Newtons second law sumF→ = ma→. -) Equation from statics, sumF→ = 0 ; sumM→ = 0 -) Same applies to dynamics For a rigid body, you have sumF→ = ma→ and then summ→ = Igα -) FBD and KD, go hand in hand For newton's second law, FBD doesent exist without the KD. FBD is only external forces and external moments -) If you have tension in the rod, and you dont cut that rod for your system. Then that tension is an internal force. Weight, normal force, friction force, external force, external moment. Those are all external to the window. -) What belongs in the KD. Mass * acceleration in KD, no forces ; In FBD no acceleration -) EOM for pure translation. -) Kinetic diagram, only one acceleration. The acceleration along that straight line -) sumF→ = max G -) sumY↑ = may G -) We're looking for the acceleration of the mass center point g. -) whenver we do rigid bodies, everything is done with the mass center point g. mass center point g is the representative point rigid body problems -) In dynamics we have sumF→ = ma→, in sum of all the moments we have sumMg = sum(mk)g = 0 -) what is the kinetic moment? In newtons 2nd law. when we look at sum of all the force (sumF), we concentrate on the FBD. and when we look at (ma), we concentrate on the kinetic diagram. The moment we write = to, we concentrate on the kinetic diagram. -) The same thing applies for rigid bodies. For sum of all the moments about 'G', (sumMg), concentrate on the FBD. and take the moment of all of these forces with respect to 'G'. Concentrate on your FBD, and look at all the external forces, and take the moment of all of those external forces. And all of those external moments with respect to the mass centerpoint 'G' -) The moment you write down = to, focus on the kinetic diagram. -) The KD, has one kinetic force mass * acceleration. -) Pure translation rectilinear, moving along the straight line. -) What happens when you have pure translation, except now you're moving on a curve. If youre moving along a curve, -) FBD, (Weight, normal force, friction force, external force, external moments). That's all the FBD. -) In the KD, how many accelerations do we have. If we are moving along a curve. We have two accelerations. we have mass* the normal acceleration, and then mass * the tangential acceleration. -) What coordinate frame would you use, if you had a curve. you can use an n-t -) left hand side must match the right hand side, must have the same convention

15.1 - 15.2 Principle of linear Impulse and Momentum

Check List*) 1. F, v, t → Impulse momentum principle For all of our problem's, we're only going to be looking at the Momentum Impulse principle of the x-y coordinate frame 2. Where is impulse taking place 3. FBD at point where impulse takes place What do you need for Momentum-Impulse*) 1.) Coordinate system First thing you're always going to need is coordinate system. 2.) The FBD at the point where Impulse/ or contact takes place Contact meaning, The force contacting the rigid body or particle. 3.) Velocity right before impact (Momentum Diagram) 4.) Velocity right after impact (Momentum Diagram) 5.) FBD → Impulse Diagram From the FBD, you get your Impulse diagram What is your Impulse diagram? Your Impulse diagram is nothing but the FBD with every force in your FBD, and Integral over that force. Remember, Impulse = Integral Fdt. You take your FBD, and you put integrals on it, that gives you your Impulse diagram. 6.) Impulse-Momentum Diagram (x-y) 7.) Impulse-Momentum Equations Tricks*) -) Problems that have to do with a Force acting on a particle over time (MIT). Momentum Impulse Time (MIT). If you have a force acting on a particle, over a certain amount of time? Think of MIT -) Problems with Force, Velocity, and Time. Use Impulse-momentum principle When you have force, velocity, or time. Think of using momentum-Impulse principle. (mv1 + sum Integral (t1 t2) Fdt = mv2) Hints*) -) Linear momentum impulse -) What is linear motion? straight line, linear motion -) What is Momentum? Momentum of any particle is the product of mass * velocity. (m*v) Mass multiplied by velocity is Momentum. -) What is Impulse? If you take the integral of the force over time (Integral fdt). That's an impulse -) How is momentum and Impulse connected with each other? -) If you're problem has distance in it, think of using WED -) WED can be simplified to COE (Conservation of Energy). When can WED be simplified to COE? No friction, and no external forces -) That's when WED becomes COE -) WED (Work Energy Distance) -) Impulse momentum problems (MIT) -) MIT stands for Momentum Impulse Time If you have a force acting on a particle, over a certain amount of time? Think of MIT -) Remember in Work Energy Distance, it was force acting on a particle over a distance. -) For MIT, distance is no longer their. It is time based -) When you have force, velocity, or time. Think of using momentum-Impulse principle. (mv1 + sum Integral Fdt = mv2) -) State 1 and State 2, In the work-energy equation, what were those two states that you had? State 1 was initial, and state 2 was the final -) In Momentum Impulse, the states are already pre defined for you. In state 1 (T1), its just before the force acting on the particle. In state 2 (T2), its the end of the force acting on the particle. -) The force is acting on a particle, from T1 to T2 -) It is super critical that you understand this -) That is the biggest difference between Momentum Impulse and Work Energy. -) The mother of all equations in dynamics is, Sum of all forces = mass times acceleration (sumF→ = ma→). -) Remember how the work energy principle is derived from (sumF→ = ma→) ? -) We're going to derive Momentum impulse, from the same equation sumF→ = ma→ -) We have sumF = ma, the a = dv/dt ; sumF = ma = dv/dt If I bring the dt, on to the other side, what do I have? sumF dt = mdv, now I have d and d on both sides, now I can integrate. Integral sumF dt = Integral mdv On the left, I could integrate with time, because I have dt. On the right since I have v, Im going to integrate over velocity. Integral (t1 to t2) sumF dt = Integral (v1 to v2) mdv. T1 and V1 match, T2 and V2 match. V1 is the velocity at time T1 ; V2 is the velocity at time T2 -) Remember T1 and T2 are the beginning of the end of time, that the force is acting on the particle. -) Here is The principle of linear momentum at moments -) Linear Impulse→ = Integral (t1 to t2) Fdt→ -) Linear Momentum = mv1→ -) From the sumF→ = ma→ We have the work energy equation. Work Energy (WED). 'D', distance is the key their We also have the Momentum Impulse (MIT). The key their is 'T' Time -) If you see a problem with distance in it, think of using the Work Energy. -) If you see a problem with time in it, think of using MIT -) Impulse momentum problems (MIT) -) MIT stands for Momentum Impulse Time -) Velocity and Force are both vectors. -) Vector equation means you have x component, y component and the z component -) Impact -) ("If you have ever played golf, their is a force acting on the ball for a short duration of time") -) As long as that golf club is in contact with that ball, the golf club is putting force on that ball. -) The moment, the ball leaves that club. Their is no longer any interaction between the club and the ball. Therefore, their is no more force acting on it. The golf club is putting a force on the golf ball for a short duration of time. -) What kind of motion do we have once the ball leaves the club? Projectile motion. -) During the time that the club is in contact with the ball, that's where we have force acting on the ball. That's where we have force acting on the ball on the particle, over a duration of time. -) That duration of time for this example could be, maybe 50 mls, or 10 mls, It doesent have to be long -) The momentum impulse principle is only valid for that short duration of time. When the force is acting on that particle. -) The moment the ball leaves the club, The golf club is no longer putting a force on the ball. -) The same principle applies on a hammer and stake When you hit a stake into the ground, the hammer is in contact with the stake for a short duration of time. -) Do you see the difference between a time based problem, and a distance based problem? -) Momentum at time t1 + Sum of all the forces from T1 to T2 is = the linear momentum in state two. (mv1→ + sum Integral (t1 to t2) Fdt→ = mv2→) -) FBD is the central topic, to any problem in this class. -) Specifically practice the FBD, Take a problem, sketch the FBD and check the solution. -) If you get the FBD incorrect, then everything else is incorrect. -) The Impulse of weight and normal force can be neglected when you're using the principle of Impulse & Momentum -) Remember, on the FBD put an Integral on every force that you have -) The internal impulses acting on a system of particles always sum to zero. -) If an impulse-momentum analysis is considered during the very short time of interaction, as shown in the picture, weight is a/an non-impulsive force. (It is non-impulsive since the weight will have a negligible effect on the change in momentum)

15.4 Impact

Check List*) 1. Line of Impact (x-axis) (y-axis 90 degrees to x axis) Identify the line of impact which is your x axis. 90 degrees to the position of your x axis, is your y axis 2. Central or Oblique. weather its central or oblique really doesent matter for the analysis. its a good idea to be able to identify something that's central or oblique. 3. COR→ BAAB21 (A hits B) & Conservation of of Linear momentum along line of impact (x-axis) along the x-axis you have two equations. COR & COM 4. If Wall/ Ground only, use (COR) Coefficient of Restitution along x-axis If you have a wall or a ground in your problem, or if you have something that doesent move before or after. Only use the COR. 5. Along y-axis : VA1y = VA2y ; VB1y = VB2y The velocity of A along the y is the same before and after. Tricks to solve Impact problems*) 1.) Identify plane of contact (Tangential to both bodies, going through point of contact) Identify the plane of contact, the plane of contact is really a tangent to both particles. 2.) LOI 90 degrees to plane of contact The line of Impact is 90 degrees to the plane of contact. 3.) LOI = x-axis & y-axis 90 degrees to-x axis The Line of Impact is the x-axis, and the y-axis is 90 degrees to the x-axis. x and y always have to be 90 degrees to each other 4.) Along LOI (x-axis): Central/ Oblique Along the x-axis, weather its central impact or oblique impact. Two rules apply. The Conservation of Momentum (COM), and the Coefficient of Restitution (COR) Along the x-axis you have two equations Conservation of Momentum (COM), and Coefficient of Restitution (COR) Regardless of what impact you have, two equations for the line of impact. two equations for the x-axis -) Along LOI (x-axis): Ground/ Wall Along the x axis if one of the particles is the ground or the wall where you have zero velocity before the impact, and zero velocity after the impact. The ground doesent move after impact, the wall doesent move after impact. If that is the case. Then you only have one equation, The Coefficient of Restitution (COR) equation. You do not have conservation of momentum at that point. 5.) Coefficient of Restitution (COR) when you drop a ball on the ground, that ball is not going to bounce back to it's original height. Some of the energy was lost during the contact. otherwise, that ball would come back straight into your hands when you dropped it. That loss of energy is described by the coefficent of restitiution 6.) Along Y-axis -) Velocity before and after the impact is the same The Velocity along y before the impact, is equal to the velocity along y after the impact. The y velocities are preserved. Hints*) -) Impact -) If you have Impact, contact, collision. Think about Conservation of Momentum. -) Remember, Impact is when two bodies come into contact with each other. -) If it's one body, and their's time involved. Think of MIT (Momentum Impulse Time). -) If you put the two bodies together, its Conservation of Momentum (COM) -) Impact is two bodies, or multiple bodies. That come into contact with each other over a short duration of time -) Think about kicking a ball, a golf ball and golf club -) The plane of contact The plane of contact is essentialy a plane that is tangent to both rigid bodies -) If you take the tangent of both rigid bodies, you get the plane of contact. -) The plane of contact can also be looked at as the plane on which both particles come together. -) The plane of contact, and 90 degrees to that, we have the line of Impact -) Always make your line of Impact your x-axis. Make your common tangent, your y-axis. -) We're only going to be looking at impact of the x and y your line of impact is going to be 90 degrees to that common tangnet. -) The y-axis is 90 degrees to the x-axis -) Central Impact The velocities before impact are in a straight line. In a straight line along the line of Impact. This would be central Impact -) Oblique Impact Where you have an Oblique angle between the velocities right before impact. right before impact you have an angle -) When a tennis ball hits the ground, thats an impact. When the ball is kicked and it hits the ground again, thats an impact -) Any time you have contact, that's an impact -) Remember, for dependent motion problems, the challenging part was the rope equations. Once you had the rope equations you could figure everything else out from their. But if you had the wrong rope equation, then everything else is incorrect. -) In this Impact problem, the challanging part is. The line of impact x and y. Where do you define your x axis and where do you define your y axis . -) If you look at a ball hitting the wall, the wall is the plane of contact. Remember, 90 degrees of the plane of contact is the Line of Impact. -) Your line of impact, by definition is the x axis. The plane of contact is the y axis. -) What is coefficent of restitution when you drop a ball on the ground, that ball is not going to bounce back to it's original height. Some of the energy was lost during the contact otherwise, that ball would come back straight into your hands when you dropped it. -) That loss of energy is described by the coefficent of restitiution (COR) -) Go to your P&F problems, and just Identify the X & Y. Practice, because if you get your X & Y swapped. Your entire calculations will be incorrect. -) Analysis of central Impact. -) 90 degrees to the common tangent, the line of impact is the y-axis -) The moment you see the x-axis, you have two equations. COM and COR. Along the x-axis you have two equations. -) unless, one of those particles is the ground or the wall. In that case, we only have one equation and that is COR -) Along the y-axis you just have COM -) Along the line of impact, conservation of momentum. Momentum is conserved along the x axis. Linear momentum is conserved -) The total momentum right before impact is = the total momentum right after. Momentum is conserved -) BAAB21, coefficent of resitution equation In BAAB21, A hits B. If A hits B its BAAB21, How does this work? ("BAAB") is ("2211") e = (VB2x - VA2x)/ (VA1x - VB1x) -) The COR is a ratio of the difference in the velocities, after impact. To a difference of velocities before impact. -) The higher the value of 'e' is , the higher the value will rise. The lower the value of 'e' is, the lower it will rise -) Line of impact is your x-axis, plane of contact is your y-axis

12.8 Polar and Cylindrical Coordinates (r, theta dot, z)

Check List*) 1.) If (r - theta - z) coordinate frame. When do you use polar coordinates for a problem? If it's a curve problem you can use the n-t coordinate system as well. When do you use the (r - theta - z) polar coordinates. When r is given, theta is given, r dot is given, theta dot is given, and theta double dot may be given. That's code for you to look at the r-theta coordinate frame 2.) Have proper book keeping (r, r dot, theta, theta dot, theta double dot ) Trick to Solving problems using polar coordinates is proper book keeping. Meaning r, r dot, theta, theta dot, theta double dot. All you have to do is plug in the appropriate derivatives, and second derivatives in the appropriate places. Hints*) -)Remember, If you have a curved path you have three choices? The n-t coordinate frame. The (x-y) coordinate frame (Projectile motion). and polar coordinates -) You can use the n-t coordinate frame for projectile motion as well. But typically for projectile motion you would use the x-y coordinate frame. The acceleration along the motion is zero (ax = 0). -)The acceleration along the y-axis is negative g, (ag = -g) if the y-axis is pointing upwards. -) Remember, regardless of what coord. frame you choose, what is the direction of the velocity vector? Tangential to the path. regardless of what coordinate frame you choose, velocity is always tangential to the path. v→ : tangential to the path. -) Regardless of what coordinate frame you choose, the acceleration vector is tangential and normal -) Tangential acceleration is in the direction of the tangent. Normal acceleration is pointing towards the center of curviture. -) For acceleration (a→), you have (an→) and (at→). -) (at→) in the velocity, since the velocity by definition is tangential. at and vt (vt→), are in the same direction, or in the opposite direction. -) Change in velocity of speed is the tangential acceleration, And at = dv/dt. -) an = v^2 / rho, (v^2 over rho, where rho is the radius of curviture). -) For polar coordinates we have a fixed origin, the polar coordinates are not like the n-t coordinate frame. The n-t coordinate frame moves with the particle. -) The polar coordinates and the rectangular coordinates have a fixed origin. In polar coordinates if you want to define the position of a particle in polar coordinates. You need two pieces of information if you want to define the position of a particle in a plane. -) For polar coordinates you need the radial distance r from the origin. And the angle that a straight line makes from the origin to the point of interest to the particle. Measured counter-clockwise from the horizontal x-axis. -) Regardless of coordinate frame the velocity vector is always tangential to the path. (use r, theta, z coordinates) -) The radial component of the tangential velocity vector is (vr = r dot). The theta component of the tangential velocity vector is (vr = rthetadot) -) The accleration vector regardless of coordinate frame will always have a tangential component and a normal component. either one of those could be zero. In general it will always have a tangential and a normal. -) The normal acceleration has a radial component and a theta component. -) This is radial acceleration and theta acceleration in the rtheta coordinate frame -) ("If you've ever sat in a twisting slide, you're moving along a curve. But you're also going down. So you have r-theta, you're turning your angle and the angle is changing. And you also have motion along the z axis.") -)What is the derivative of the r-component, what is the r for the person going down the slide. What is r measured from the column? r is basically the radius. What is r dot? (Is the person moving out radially? no.) This person is basically on a circle as he goes down. it is going to be zero, r double dot is also going to be zero. Is theta dot also zero? no, theta dot is changing, theta double dot is also changing. -) Speed in the polar coordinates is the sum of the square root, of the sum of the squares in the individual components. Just like we had the speed in the x-y, as (vx^2) + (vy^2). Here, we have vr^2 + vtheta^2 + (If you have a z component) vz^2. Speed = sqrt( vr^2 + vtheta^2 + vz^2) -) In a Polar coordinate system, the Velocity vector can be written as v = vr ur + vthetautheta = rur + rthetautheta. The term theta is called Angular velocity -) The Speed of a particle in a Cylindrical coordinate system is Sqrt ((rthetadot)2 + (rdot)2 + (zdot)2) P.s*)

13.1 - 13.3 Equations of Motion

Check List*) 1.) Newton's second law is a "law of nature"-- experimentally proven, not the result of an analytical proof. 2.) Mass (property of an object) is a measure of the resistance to a change in velocity (inertia) of the object. 3.) Weight (a force) depends on the local gravitational field. Calculating the weight of an object is an application of F = m a, i.e., W = m g. 4.) Unbalanced forces cause the acceleration of objects. This condition is fundamental to all dynamics problems! Procedure for the application of the Equation of Motion*) 1.) Select a convenient inertial coordinate system. Rectangular, normal/tangential, or cylindrical coordinates may be used. 2.) Draw a free-body diagram showing all external forces applied to the particle. Resolve forces into their appropriate components. 3.) Draw the kinetic diagram, showing the particle's inertial force, ma. Resolve this vector into its appropriate components. 4.) Apply the equations of motion in their scalar component form and solve these equations for the unknowns. 5.) It may be necessary to apply the proper kinematic relations to generate additional equations. Hints*) -) The forces applied on Free Body diagram's are spring forces, friction forces, reaction forces, external forces, and weight. -) Their are no internal forces on a Free Body diagram -) Kinetic diagram is Kinetic Force : ma→ -) Newton's second law is sumF→ = ma→ (Sum of forces equals mass time acceleration) -) sumF→ is the FBD -) ma→ is the KD -) Free body diagram is external forces -) Kinetic diagram is the ma→ -) The accleration is super important, because the acceleration of an object is dependent on two things. The mass of the object and the force. The net force applied on the object. The greater the force, the higher is the acceleration. The smaller the mass, the higher the accleeration. -) The free body diagram contains different forces, the friction force, the weight, the reaction forces, spring forces, external forces -) Kinetic force is Mass multiplied by acceleration. sumF→ = ma→ (Sum of all forces equals to mass time's acceleration) -) For free body diagram, external forces -) For kinetic diagram, look at the accleration. Multiply the accleration with the mass. That is your kinetic diagram -)Example -) You have a block, you have a force being applied on that block, and you have the block compressing the spring. The spring is attached to the wall -) The free body diagram of the system, how many elements do I have in this system? I have a spring, I have a block, and I have this rod. -) I can draw a free body diagram for either one of those. Or I can draw a free body diagram for the entire system. -) Remember, for the free body diagram look at all of the external forces -) If the entire particle is my system. I have the weight of the block (W), I have the external force (F) (F = 500N), I have a friction foce (Ff), I also have a spring force (Fs) -) Do I really have a spring force? If this is my entire system? Remember, the spring is going to exert a force on the object. The object is exerting an equal and opposite force as well. -) If I look at is as one system, those two forces cancel each other out. -) For the kinetic diagram, I have to look at element by element -) For the block, the block has a mass, does it have an acceleration? is the block moving? Yes, the block is moving. -) What direction is the acceleration of the block? Its moving to the right. -) Remember the kinetic diagram is just mass multiplied by the acceleration. -) Does the block have any acceleration in the vertical direction? No -) What about this rod, does thos rod have acceleration? Is the rod moving at all? yes, the rod is connected to the block so its gonna move. -) The kinetic force of a rod? mass multiplied by the acceleration. -) Im going to assume that the rod is mass-less, no mass. therefore, no kinetic forces for the rod -) what about the spring, does the spring move? Yes the spring moves, it has an acceleration. But im going to assume the mass of the spring is zero. therefore no kinetic force. -) I only have one kinetic force for this system. which is mass times acceleration. (sumF→ = ma→) -) we have three parts to this system, we have the block, we have the rod, and we have the spring. -) If I wanted the free body diagram of just the rod, the rod has no mass, Mrod = 0 ; Mspring = 0 -) If I just wanted the free body diagram of the rod. What do I have? I have the external force F. I have no weight. I also have the reaction force of the rod, when it is connected to the block. -) Remember I seperated the rod from the block, so I have to take into consideration the reaction forces. -) For reaction forces , we have point A, Ax, and Ay (point A pointing in the x and y directions). -) What does the kinetic diagram of this rod look like? -) Remember the trick, look at the acceleration of the rod. Multiply that by the mass of the rod. The acceleration of the rod is A, the mass of the rod is zero. This rod, Mass-less, It has no kinetic diagram. -) The sumF→ is equal to zero for the rod. -) Lets look at the block? For the block, I have the weight of the block (Wb), I have the normal force of the block (N), we have the friction force (Ff), We have the external force (F). We have the reaction forces (Ax), and (Ay). -) If im seperating the block from the rod, I will have reaction force. -) Now, we also have the spring force (Fs). -) The free body diagram changes based on where you're system is. -) If you look at the entire thing, then you have one Free body diagram. But if you're looking at individual components then you have to worry about reaction forces. -) The reaction forces on A are all internal to the system. They all cancel each other out. Because Ax will cancel out this Ax, and Ay will cancel out this Ay. When it is put together. -) What does the kinetic diagram of the block look like? The block has a mass, and the acceleration off the block is ma→ -) Example -) A simple pendulum, the free body diagram of this pendulum. Lets assume that this rod has negligible mass. All the mass is consentrated on that pendulum bar -) The free body diagram of this pendulum is going to look like what? You have the weight of the pendulum. no weight for the rod because im assuming it has no mass. Im going to seperate the pendulum from the base, If I seperate the pendulum from the base, will I have reaction forces? Yes, I will have Ox and Oy -) What is my system here, for this free body diagram. The rod with the pendulum bar. -) If I made the entire thing my system, meaning I'm not going to seperate the rod from the support. What happens to the FBD? The Ox and the Oy cancel each other out. -) When I seperate the rod from the base, It has reaction forces. But if I look at the base itself, just the base. What does the free body diagram of just the base look like? Ox and Oy, when I put them together Ox and Oy are going to cancel each other out. If i look at the entire system. The base and the pendulum as my system. -) This is the free body diagram of pendulum only. This is the free body diagram of base. -) If I made the entire particle my system. Then I would only have the weight (Wp) of the pendulum -) It all depends on where you define your system -) If I wanted to determine the tension, the bob has a weight. Therefore that weight is going to pull on that rod. Imagine that rod is a string. That bob is going to pull on that string. -) If I asked for the tension on that string, where would you put your system? In the middle of the rod -) FInd tension in string, then my system becomes this. The free body diagram of my system is going to be the weight of the pulley and the tention force. Because when I cut that string, Im going to have a tension force. -) What does the kinetic diagram look like? -) The take away is, the free body diagram depends on where you define you're system -) The kinetic diagram, look at the mass. And look at the acceleration of that mass. -) Is the mass moving along striaght line? along two directions? or along a curve? Its moving along a curve. -) If it moves along a curve, how many accelerations does that barb have. It has two accelerations. -) When looking at the barb, it has two accelerations. Normal (Man) and tangential (Mat). -) Man, and the tangential (Mat) is pointing towards the center of curvature. -) For you're kinetic diagram, all you do is take the mass and mulitply it by the acceleration. -) What is the kinetic diagram of this mass'les string? Mass of the string is zero, so the kinetic diagram is zero. -) Example -) You have a pendulum, and the pendulum is attached to a spring. Assume that it is Mass-les. -) If this is my system, then the free body diagram I have is (Wp) weight of the pendulum. -) Do I have reaction forces at O? Yes, because Ive seperated the string from the support. Ox, Oy. -) I also have a spring force -) ("Right now, I am standing on the floor. I'm pushing against the floor, the floor is pushing back against me. For my FBD, If I seperated myself from the floor. It would be Mg and N ") -) The kinetic diagram of this system, we would use Mat and Man. Because everything else has no mass, the spring has no mass. The string has no mass -) Example (Dependent problem) -) How would I determine the tension in the string? I would make my system. cut it right at the string. Then for B, since I need the tension in the string. I would do the same thing, cut it right at the string. -) What does the FBD of my first system (A) look like? Weight of the block (Wa), the normal force (N), Friction force (Ff), and the tension force (T). -) What does the kinetic diagram look like? The acceleration of the mass m is only in the horizontal direction. maA (mass acceleration of A). -) What does the FBD of B look like? you have the weight (Wb). How many strings are attached to 'B'? Two. How many tension forces am I going to have? Two. -) Remember, tension force is the same throughout the entire string. Otherwise you'd have a slack -) What does the kinetic diagram of B look like? If acceleration of A is going to the right. The acceleration of B is going down (mBaB). -) Now I can apply newtons second law, I will place an X and Y axis. -) For A (Sum of all forces along the x-axis) What forces do I have for the X-axis? For A: X : T - Ff = mAaA Y : N - wA = 0 (you only have one kinetic force on X direction) (In the y-direction you have normal force and weight) (The moment you write = to, you can cover up the FBD and only look at the KD) (Look at KD, and Kinetic forces along the X-direction) For B: Y: 2T - wB = -mBaB -) Example You have two particle's mA and mB, one particle is pushed to the right by an external force P -) If I draw my system like this, what is the FBD of the mass A? Weight of a (Wa), Normal force on a (Na), The string force (Fs), and the external force P -) What does the kinetic diagram look like? For the kinetic diagram, only look at the acceleration. Is A moving? yes it is. Its going to move in the right. Mass of A multiplied by the acceleration (mAaA) -) Newton's second law can be written in mathematical form as sumF = ma. Within the summation of forces, sumF, ________ are(is) not included? Internal forces -) The equation of motion for a system of n-particles can be written as sumFi = sum mi ai = maG, where aG indicates Acceleration of the center of mass of the system

Exam 3 Review

Check List*) Process to solve a Kinetics Rigid body problem 1.) Draw a Coordinate Frame First thing is to draw a coordinate frame 2.) FBD & KD (Direction of Friction Force) 3.) sumF → = m aG→ sumMG = IGα sumMp = (sumMk)p 4.) Round Rigid Bodies (Rolling or slipping), (Direction of Friction Force (Opposes action)) Tricks*) Friction will oppose Moment about Mass center of Gravity (Round object). If Moment about "G" is zero, friction will oppose Driving Force. Important Equations*) 1.) omega = [0 0 - omega] ; α = [0 0 α] 2.) Va = omega x rA/o |Va| = omega(3) 3.) aA = aA*t + aA*n = α x rA/o - omega^2 rA/o 4.) vB = vA + vB/A vB = vA + omegaAB x rB/A 5.) aB = aA + aB/A aB = aA + α x rB/A - omega^2 rB/A 6.) For acceleration, first velocity analysis 7.) Think vectors Hints*) -) Kinematics is the study of position, velocity, and acceleration -) The kinetic diagram had ma, and IGα Kinetics is the study of forces, moments, and acceleration, linear acceleration or rotational acceleration. -) Can you look at a problem, and can you identify what kind of motion each rigid body in that problem is going through. -) look at every problem and ask yourself, what kind of motion does this go through? -) Based on the type of motion, the analysis is completely different. -) If you have translation, remember their are two types of translation. Rectilinear translation and curvilinear translation. Make sure you can recognize both -) If you have translation, remember absolute motion analysis. so you have an absolute coordinate frame You define the position of that rigid body with respect to theta, the angle, or whatever -) The first derivative of position gives you velocity, second derivative of position gives you acceleration -) The chain rule, the product rule -) Rotation -) When you see a rotation problem, the first thing you need to ask yourself is, what is the point of rotation. The axis of rotation is always normal to the paper, coming in and out of the paper. We only have one axis of rotation -) Which point is the rigid body rotating in -) You need to be able to identify if it's rotating about a mass center point 'G', or a mass center point 'O'. -) For rotation, you have constant rotational acceleration equations, and variable rotational acceleration equations (Rigid body motion about fixed axis Eq. sheet) -) Just as you did for translation, constant acceleration and variable acceleration equations. -) You have the same type of equations for pure rotation -) General Motion -) General motion is the addition of translation and rotation. -) You should be able to look at a rigid body and determine if that rigid bodies going thorugh rotation, translation, or general motion. -) General motion is relative motion. You dont have relative motion when you have pure translation and when you have pure rotation. -) You have relative motion when you have general motion Their are two types of relative motions in general motion, one is the relative velocity analysis -) In the relative velocity analysis, the velocity of 'B', A and B are two points on the same rigid body going through general motion -) The velocity of B is equal to the velocity of A + the relative velocity. Vb = Va + Vb/a (Remember the trick for this equation) -) Vb/a is equal to omegaab, omegaab is the rotational velocity of that link, that's going through general motion. omegaab cross r B/A (omegaab x r B/A) -) ( Vb/a = omegaab x r B/A ) r B/A ; It starts at A and ends at B, starts at the bottom and ends at the top. -) Remember, the acceleration always has two components. Normal and tangential -) aB = aA + aB/A. The same trick applies, a and a cancel out, we are left with aB = aB -) aB with respect to A, aB/A, two components. Acceleration always has two components. -) α cross b with respect to a. αB/A = α AB x r B/A - omega^2 (AB) * (rB/A) -) General motion is also pure rotation about the instantaneous center of rotation. It only applies for velocity analysis, not acceleration analysis. -) If omega is clockwise, then the vector is negative. -) Kinetics -) If the problem asks for a force, or an acceleration. Think of using the EOM. FBD and KD -) For EOM, the first thing you need to think is what kind of motion do I have? pure translation? rectilinear or curvilinear? -) If you have translation only, translation only means no rotation. you only have these two equations, sumFx = m(ag)x , the accleration of the mass center point g The same thing for y sumFy = m(ag)y -) The moment you go into rotation, you have to differentiate. Is it rotation about the mass centerpoint 'G', or is it rotation about a different point. Based on that your EOM is going to change -) If you have rotation about the mass centerpoint 'G', the acceleration of the mass center point 'G' is zero It's not moving, its fixed. Because your axis is going right through 'G' -) The sum of all the forces sumF = Mag, which is zero. The sum of all the forces in the y, sumF = Magy, is zero. remember (ag=0) -) Because of rotation you have Igα in your KD sum of all the moments about 'G', is Igα. -) If you wanna take the summ of a different point, then its summp -) FBD, take the sum of all moments about that point p = to Kinetic diagram, sum of all the kinetic moments about p. sumMp = sum(Mk)*P -) If you're rotating about a different point 'o', then the mass center point g is going to have an acceleration. -) If you have general motion. EOM general motion -) sumFx = m(ag)x ; sumFy = m(ag)y sumMg = Igα or sumMp = sum(Mk)p -) If youre given Kg, kg → IG = mkG^2 If youre given Ko, Ko → Io = mKO^2 -) The moment of inertia is mass * the radius of gyration^2 -) The parallel axis theorum, I with respect to any point on that rigid body is equal to Ig + total mass* normal distance that points to g^2 ( I = Ig + md^2 ) -) Equations of motion for round objects. For round objects, the first thing you ask yourself is. Rolling or slipping? Is that round object rolling or is it slipping? -) What is the direction of the friction force, for a round object? Friction force opposes the action. -) Look for the moment at the center of that wheel, the friction force is going to counter that moment. -) If you are rolling, the Ff is unknown. It can be zero, it can also be greater than zero. The maximum value it can take is MSN -) If your sliding then you dont have a kinematic constraint -) If youre sliding, then the friction force is going to be in the opposite direction. -) Ff = Msn (slip impending) Ff = Mkn (slipping) -) What is the maximum force I can apply, before the wheels slip?

17.4 Equations of Motion (Rotation about a Fixed Axis)

Procedure for pure rotation problems*) When do you use EOM, If the problem is asking for a force or if the problem is asking for an acceleration. You use the FBD, KD EOM 1.) Rotation about G or some other point Is the rotation happening with respect to the mass center point 'G', or is it happening with respect to some other point. -) If its happening with respect to 'G', then no Magn and no Magt. Only Igα. If its happening with a different point other than the mass centerpoint G, then you have Magn, Magt, and Igα 2.) Rotation about G → KD only has IGα 3.) Rotation not about G → KD has IGα & Magn + Magt -) Do not forget IGα Make sure to never forget the Igα 4.) coord frame (n-t) 5.) FBD 6.) KD 7.) EOM : n, t, sumM ; FBD = Kd 8.) Compute IG or IO If you're taking the sum of all moments with respect to the mass centerpoint 'G', or if you are taking it with respect to a different point, O, P ,Q. You would take the moment of Inertia about that point. 9.) 3 equations, 3 unknowns Their are 3 EQ. and 3 unkowns 10.) Look for Kinematic constraints Kinematic constraints are basically where you have a round object. Where you have a round object rolling on a flat surface Then the velocity of the mass centerpoint 'G' is omega of the rolling object * the radius. The tangential accleration is alpha * the radius r. Hints*) -) last class we did EOM : pure translation -) Your FBD has all kind of forces, weight, normal force, friction force, external force. You can have moments on your FBD, external moments -)The KD is only going to have two kinetic forces Max, May. If its curvilinear then Man and Mat. Their is no alpha, because alpha is zero when you have pure translation or curvilinear translation -) Pure rotation Pure rotation to this kinetic diagram, you are going to add one more kinetic moment, (Igα). Ig is the mass moment of intertia of that rigid body with respect to that mass centerpoint G. α is the rotational acceleration, you are going to add Igα. -) For any rotation problem, the first thing you are going to ask yourself is. Is the rotation to a mass centerpoint G, or is the rotation with respect to a different point other than mass centerpoint G. -) If you are rotating about G, what is the most important point of the rigid body from a dynamics point of view? The mass center of gravity 'G'. -) If you are rotating the mass center point with respect to gravity, what is the acceleration of the mass center of gravity? zero. -) If the acceleration of the mass center of gravity is zero. do I have an Magx and an Magy, or an Magn and an Magt? No But we will always have an Igα. -) If you have pure rotation about the mass center of gravity 'G'. Then your KD is going to look like, 'G' your center of gravity. and Igα -) magx = 0, magy = 0 -) Their are only two cases of pure rotation. One is rotation about the mass center point 'G', the other is rotation about a different point other than the mass centerpoint 'G'. -) The equations of motion (EOM), -) What do the EOM look like? The EOM is the axis of rotation, the rotation axis. Is it going through 'G', the mass center point 'G'? or is it going through a different point other than the mass center point 'G'? -) That's the first thing you have to define? That's the first thing you have to answer when you are doing a rotation problem. Is it rotating about 'G', or is it rotating about a different point? If it is rotating about a different point, then the normal acceleration and the tangential acceleration. The mass center point 'G', is going to have a normal acceleration and a tangential acceleration. Why? Because 'G' rotates about that point 'O' on a circle. -) If you have a vertical rod, rotating about a a hinge at point 'O'. It is pure rotation -) If you have pure rotation, the first thing you need to ask yourself is. -) Is the rotation about the mass center point 'G'. Or is the rotation about a different point.

16.6 Instantaneous Center of Zero Velocity

Tricks*) -) Any general motion problem can be converted into an instananeous center rotation You need 3 quantities to solve the general motion of any problem. Hints*) -) Their are three different types of motion. Pure translation (rectilinear, curvilinear), Pure rotation, and general motion (combination of translation and rotation) -) If you have a link that's changing in angle, and its moving. Then that link is in general motion -) If you have a link that's fixed at one end, and its rotating. Then its not translating, That's pure rotation. -) For general motion, any two points on that rigid link with general motion. Have different velocities. -) Have different velocities and one omega -) The definition of rA/B is a vector that starts at B and ends at A. whatever you're taking with respect to is the starting point and the end point is the top 'A'. -) What is general motion? relative motion. -) For relative motion analysis, you need three quantities. Velocity, direction, magnitude. -) Relative velocity equation VA = VB + VA/B = VB + omegaAB x rA/B VA = to VB plus the relative velocity of a with respect to B The relative velocity of A with respect to B is omega of that link cross rA/B. What you have on the left, must match to what you have on the right -) Pure rotation The velocity of every point on that rigid link, pure rotation is 90 degrees to that link If you have pure rotation, that velocity is always 90 degrees. Any point will have a velocity 90 degrees to that link. Va, velocity is just omega * the length of the distance of that point to the center of rotation -) Instanteous center of rotation. At that particular instant, the center of rotation is their. At another instant, the center of rotation is going to change. It's only valid at that particular instant. -*) We have two methods to solve a general motion problem -)Relative velocity analysis Va = Vb + omegaab x ra/b -) Instantaneous center of rotation Any general motion problem is a pure rotation problem about the instantaneous center of rotation. -*) Characteristics of IC -) The instantaneous center of rotation may or may not be on the rigid body. It doesent have to be on the rigid body -) The center of rotation always has zero velocity -) Instananeous center of rotation is only valid for that particular instant in time. As time changes, that IC is going to change. -) Instantaneous center of rotation is only valid for an instant in time

17.5 Equations of Motion (General Plane Motion)

Tricks*) 1.) Friction always opposes action Action is what causes that round object to either roll or slide. If I have a free rolling tire, what causes that tire to rotate. If its free rolling omega, the rotational speed causes that tire to rotate. If omega is the action, then the Ff is going to oppose that action. The friction force is going to create a moment to slow omega down. The friction force always opposes the action. We want the friction force to be as low as possible. 2.) If problem gives you Ms & Mk then check rolling vs. slipping 3.) If problem gives you Ms only, rolling Ff unknown If the problem only gives you MS, and no MK. Then the tire is rolling. 4.) If problem gives you Mk only, slipping Ff = Mkn If the problem only gives you MK, then you know that the tire is slipping. 5.) If problem asks for max value, Ff = MsN If the problem asks for Max force, or Max torque, then you can assume the Ff is = to MSN. Solution process*) 1.) Assume rolling : ag = rα If you assume rolling then you know that ag = rα. That's the kinematic constraint, that's your fourth equation. four equations, four unknowns 2.) Solve for F Solve for the friction force F 3.) Check rolling assumption : F < MSN When you are rolling, the friction force must be less than MS * N (MSN). The moment the friction force is greater than MSN. That means the static friction coeficent is broken. so now it is sliding. 4.) If F < MSN, then rolling assumption correct If the calculated friction force is less than MSN, you are rolling and the assumption is correct. 5.) If F > MSN, then slipping has occured If the friction force you calculated, is greater than MSN. Then slipping has occured. -) If slipping has occured, then your assumption of ag = rα is incorrect. If slipping has occured, then the friction force is MKN, and you can solve for ag and alpha. If F = MKN, solve for ag and α Important points*) 1.) Round object motion, check for rolling or slipping (Ms & Mk), First assume rolling, you dont know if this round object is rolling or slipping. 2.) Apply the appropriate kinematic conditions, & check your assumption 3.) Friction force always oppose action, that causes motion 4.) General motion, KD will have Magx, Magy, and Igα. Hints*) -) We've done EOM for pure translation, curvilinear translation. And EOM for pure rotation -) When you have pure rotation, you have one more component in your KD, which is Igα. -) For rotation you have to ask yourself if it's happening with the mass centerpoint 'G', or is it happening with respect to a different point. If it's happening with respect to a different point, your KD will look a little different. -) Their are two topics for todays lecture, one is EOM for general motion, General motion is translation + rotation The KD is going to have the acceleration of the mass center point 'G', because its translation. Its also going to have the Igα from the rotation. -) The equations of motion for round objects. Round objects can be an entire wheel, a sphere, a ring When you have that in your problem, their is something you have to pay very close attention to. -) That is a round object, It can do one of these three things. It can roll, Imagine placing a round tire on an incline plane. If the angle of that incline plane is not that steep. If it's flat, then that tire will just roll down down the plane. -) What if I made the angle really steep? If the angle were really steep then whats going to happen is, the tire is gong to roll and slip down the plane. Think about a car, when you are at a traffic light. If you slowly press the gas pedal your tire is going to roll. If you continuously press the gas pedal a little harder, then the tire is going to slip. -) When you're driving on a thick sheet of ice, if you want your tire to roll on that ice. You have to drive very slowly. You have to put very little torque on that wheel for it to roll. Any amount of high torque is going to cause the friction of that tire and the ice to break, causing it to slip. Its not going to roll, its going to slip -) MuS * N (MSN) -) When you have a round object, you have to think of either rolling or slipping. Their is a very fine transition between rolling and slipping. Its called slip impending -) Based on these three conditions, your FBD and the KD dont change. What is going to change is the value of the friction force. -) When your tire is rolling, your Ff is unknown, for rolling tire the friction force is greater than zero, but less than MsN (The static coefficient), it can be anywhere inbetween. (0 < Ff < MsN) The friction force is unknown when you are rolling -) When slip is impending, the friciton force reaches the maximum value, Ms * N (MSN). Slip impending is slip about to occur -) When you have full slip, when you have full slip occuring friction force is MK * N (MKN). -) In the past problems. Friction force is = to MK * N, (Ff = MKN) The reason is, because we have assumed it is sliding. -) Sliding friction force is MKN, But if you have a round object and your rolling, its never MKN. Unless you are sliding on the surface. -) Their are two coefficients of friction, static coefficient of friction (Ms), and the kinetic coefficient of friction (Mk), Static coefficent of friction is when you're static. -) If you're trying to move this table, Im applying a force but the table is not moving. That means something is pushing against, that something is that friction force. That friction force is pushing against me, hence im not moving. The sum of all the forces is zero. But when I break that friction force (MSN), then it starts to move. You have to break that friction force, you have to break that friction contact for motion to start. That's a static object -When you're rolling, your Ff is anywhere between greater than 0, and less than MS*N (MSN). -) The moment your tire starts to slide or slip, then it becomes MKN, MuK * N -) The dynamics of a rolling tire, it starts with omega = 0. The omega starts to increase For the first few seconds the tire is going to roll, then slip is just about to happen, then you start to slip -) This is what happens when you are starting to apply a torque on the tire. -) This is what happens to your tire when you stop at a red light, and you give it full pedal position. It's going to roll, slip is about to happen and then it starts to slip. -) What happens when the tire is slipping with respect to the ground. That's when you're burning your tire -) When you have a round object, that round object can either roll or slip. -) If that round object rolls, nothing is different with the KD. the friction force is unknown, you have another equation called the Kinematic constraint. -) The kinematic constraint is the acceleration of the mass center point G. is equal to r * alpha. -) You either have the friction force, or the Kinematic constraint. You cannot have both. -) When your rolling, you have the kinematic constraint, and the Ff is something you have to compute -) When your slipping, the equation is MKN, but you dont have the Kinematic constraint anymore. -) When that tire is slipping, the tire is slipping and you are burning the tire, their is no kinematic constraint between the acceleration of that mass centerpoint of the tire, and the rotational acceleration of that tire. -) You either know the Friction force, or you have the kinematic constraint. -) When you're rolling, you dont know what the friction force is but you have the kinematic constraint. When your slipping you know what the Ff is, but you dont have the kinematic constraint. -) Pay attention to the problems, where you have a round object rolling -) If your problem has a round object, either the kinematic contraint is known? or the friction force is known? but no both. -) Rolling vs slipping? Rolling you have the Kinematic constraint, Ag is r alpha, Vg is r omega and the friciton force is unkown. The friction force is between zero and MS * N (MSN) I want to say it is less than, because if the moment is equal to MSN, then slipping is just about to happen. -) When you are slipping, the kinematic constraints are gone. But your friction force is MKN, Mu K * N -) If the problem says the slip is just about to happen then the friction force is MSN. If you know what your friction force is, then you cant assume you have the kinematic constraint as well -) The kinematic constraint for a round tire is, r* alpha = ag That's the kinematic constraint for a tire, or for any round object. -) Non round objects, you dont have to worry about since they are not rolling. Hence you dont have to worry about the rolling side, only sliding for non round objects (MKN). -) For a rigid body, you can either have a round object, or a non round object The friction force for a non round object, as we have seen before. The direction of the friction force is always going to oppose the motion. -) For a round object, its a little more tricky For a non round object, the friction force always opposes the action. the action that causes that round object to roll or slide The direction of the friction force is importnt, if you put the direction of the Ff in the wrong direction. The EOM will not give you the correct answer. Make sure you have the direction of the Ff correct. -) Action is what causes that round object to either roll or slide. If I have a free rolling tire, what causes that tire to rotate. If its free rolling omega, the rotational speed causes that tire to rotate. If omega is the action, then the Ff is going to oppose that action. The friction force is going to create a moment to slow omega down. The friction force always opposes the action. We want the friction force to be as low as possible. -) You dont know what the friction force is, if its rolling -) When you are rolling, you do not know if you are going to be rolling or slipping. The only way you can figure that out is by calculating the friction force, and then see if the friction force is less than MSN. All the way up to MSN, the tire is rolling. -) The moment that friction force is equal to MSN, slip is about to happen -) The moment that friction force is greater than MSN, It has already slipped. Then F = MKn -) If the problem asked for, what is the maximum force P, that I can apply to the disk, so that the disk rolls without slipping Remember the Ff MSN, is the maximum force. The friction force can be all the way up, the moment the friction force is = to MSN, slip is about to happen -) The maximum force I can apply, the maximum torque is when the friction force is = to MSN

13.4 Equations of Motion (Rectangular)

Tricks*) 1.) Problem asks for F or a → FBD + KD & Newtons Second law If your problem ever asks for a force, or if your problem ever asks for an acceleration. think of using the free body diagram and the kinetic diagram. 2.) If all Forces are Constant → a = c If the sum of all the forces (sumF→ = ma→) is contant, then you have a constant acceleration problem. 3.) If any Forces = f(t) → a = f(t) 4.) Motion of Particle rectilinear, use x-y coordinate frame If its motion along x and y, a projectile motion is along x and y. Is projectile motion also a motion along a curve? Yes it is. But for projectile motion, we will use the x-y coordinate frame 5.) Motion of Particle curvilinear, use n-t or r-theta coordinate frame If you're moving in a circle, the best thing is to use the n-t or r-theta coordinate. 6.) Each particle with mass, gets its own FBD and KD If that particle has a mass, it gets its own FBD and KD Remember*) 1.) FBD → only Forces, no Internal Forces 2.) KD → ma 3.) Each particle with mass gets its own FBD and KD Hints*) -) The class prior to this we looked at Newton's second law. FBD and KD, the FBD depends on where you draw you're system -) Now we want to use Newton's second law. to determine the equations of motion for a particular particle. -) What comes first, force or acceleration? Force comes first. First the force, as a result of the force the the object moves, When an object moves it has a change in positon. Thats velocity. Then theirs a change in velocity, and thats acceleration. -) Remember we have three different coordinate frames, by now you should know which coordinate frame to use with which particle. -) sum of all the forces equals mass times acceleration (sumF→ = ma→), if the sum of all forces is constant, what will the acceleration be? constant. -) Constant accleration problem? then you would know which of the three equations to use. -) If you're moving in a circle, the best thing is to use the n-t or r-theta coordinate. -) In this example, their are two particles with mass. The person doing the lifting and the stack of weights itself. You would have two FBD and KD for this problem. -) If it has mass, it has its own FBD and KD -) How would you draw the FBD of the stack of weights at B? you're system would look like this? that would be the system for the stack of weights. -)For the other mass, If the system goes through a rope, then of course their is a tension problem. If the system includes the rope? Then, their is no tension because tension becomes an internal force -) Remember and practice these two tricks. Each particle with a mass gets its own FBD and KD. For the kinetic diagram its mass times acceleration. In the kinetic diagram you only have the accleration of one mass. -) Newton's second law says sum of all the forces equals to mass times acceleration. (sumF→ = ma→) -) Sum of all the forces, look at your FBD -) The moment you write down equal to, look at your KD -) In your KD, only mass multiplied by acceleration. -) In your FBD, only the forces within your system. -) We're going to practice Newtons second law on a problem. Here we have two particles with a mass. The box The FBD of the box is going to be, you have Mg the weight of the box (Wb), you're also going to have the normal force (N). youre going to have the friction force (Ff), and the external force (F). That's the FBD of the box -) The kinetic diagram of the box, ask yourself. How many accelerations does the box have? in this case, it only has one. it is ma x, since its on the x axis. -) each particle with mass gets in own FBD and KD. If I look at the other system, again weight of this system (Wp). Their is the normal force (N), Their is a friction force (Ff) -) If the person is pushing on the box, with a force (F). The box pushes back with an equal or opposite force. -) Label the x and y. -) The kinetic diagram of this person looks like this. Mass multiplied by acceleration (Max) on the x axis. We're going to assume that its the same acceleration. -) Now lets look at the FBD of this dragster. How many particles with mass do I have in this system. I have two particles with mass in this system. we're assuming the parachute has a mass. That's valid, the parachute will have a mass. -) If its two systems, im going to draw my window around the first mass. Its going to have a center off gravity (g), the weight of the dragster. -) its going to have Normal force, friction force. Theirs a friction force at the tire. friction force F is in the front tire and the rear. we're going to assume their are different friction coefeccients. -) The force of the parachute, and the drag force of the parachute. -) what does the kinetic diagram look like? how many accelerations does the dragster have? one to the left, mad (ma dragster). -) Now your going to set your FBD equal to your KD, and you have your sum of all forces equals to mass times accleeration. -) how would we write newtons 2nd law for the dragster, sum of all forces in the x direction (positive direction of x), -) x and y, I can set my x and y to any direction I want to, as long as they're 90 degrees to each other. thats the only constraint to x and y, and n and t, although t has to be in the tangential direction and n has to be pointing towards the center of curviture. -) what forces do we have in the x direction? friction force (Ff), negative friction force. friction force in the front minus the friction force in the rear. plus the drag force, the positiive x direction. equal tp -) The moment I write equal to? I can fully focus on my KD. -) what does my kinetic diagram say, mass times acceleration. is that positive or negative? negative. - Mad (Normal accel. of dragster). sum of all the forces in the x direction. -) Sum of all the forces in the y direction, is not that interesting. because we dont have acceleration in the y. -) If the dragster had its breaks on, the tires are not rolling. If the dragster is being dragged on the ground. then, the tires are not rolling anymore, the tires are sliding. -) When its sliding, the friction force changes direction, opposite to motion. -) If its sliding, then the friction force is opposite to motion -) If its rolling, then the friction force is in the same direction. -) Newtons second law. (sumF→ = ma→) -) force is a vector on the left side, acceleration is a vector on the right side. since its vector vector. it has an x comp and y comp -) Remember, one particle one mass, one FBD one KD. -) If your problem asks for a force or acceleration, then remember FBD and KD. -) In Dynamics, the friction force acting on a moving object is always a kinetic friction -) If a particle is connected to a spring, the elastic spring force is expressed by F = ks. The "s" in this equation is the Difference between deformed length and un-deformed length

16.4 Absolute Motion Analysis

Check List*) 1. Define a fixed coordinate frame First thing you need to do, is define your fixed coordinate frame. Can a fixed coordinate frame be moving with the rigid body? No, it has to be fixed to the ground. That's why its called the absolute coordinate frame. 2. Look for a rigid body that's in pure translation Look for the rigid body that's in pure translation 3. Define position of rigid body in pure translation with respect to fixed coordinate frame. When trying to analyze a rigid body in pure translation, the first thing you need to do is fix your coordinate frame. Then look for the rigid body that's in pure translation. Then define the position of the rigid body, with respect to that fixed coordinate frame. Just like when we were doing dependent motion, we would define a datum, and then we would define the position of that particle with respect to the datum. 4. Derivative of position to velocity Take the derivative of position, the derivative of position gives you velocity. Derivative of velocity to acceleration. The derivative of velocity, that gives you acceleration. We're trying to determine the velocity and the acceleration of this rigid body thats pure translation Tricks*) 1.) Chain rule and Trigonometry We're going to use both chain rule and trigonometry. If x = L1costheta = f(theta) ; xdot = -L1sin thetadot x is a function of theta f(theta). so, dx/dt is equal to (dx/dtheta) (dtheta/dt) dx/dt = dx/dtheta dtheta/dt, the theta cancel out and you have dx/dt = dx/dt Hints*) -) Rigid body, but pure rotation about one axis. -) Rigid body pure rotation. -) Question you ask yourself when you see motion, weather its translation or rotation. Is the acceleration constant, or is it varying with time or position. -) Translation -) If you have a rigid body that's translating. Can we find the velocity and the acceleration of that rigid body that's translating. -) Is the acceleration constant, or is it varying with time. -) Absolute coordinate frame -) Absolute coordinate frame, is a fancy word for Fixed coordinate frame. -) Every dynamics problem, every statics problem. The first thing that you need to do is define your coordinate frame. -) When trying to analyze a rigid body in pure translation, the first thing you need to do is fix your coordinate frame. Then look for the rigid body that's in pure translation. Then define the position of the rigid body, with respect to that fixed coordinate frame. Just like when we were doing dependent motion, we would define a datum, and then we would define the position of that particle with respect to the datum. -) The objective for this topic, is to determine the positon, velocity, and acceleration of that rigid body going through pure translation. -) Identify rigid body with pure translation -) ABS coordinate frame -) Define Xa & Yb with respect to ABS coordinate frame. -) The first thing we want to do, is identify the rigid body in pure translation.

18.1 - 18.3 Work of a Force, Couple & Moment

Check List*) 1. FBD Look at all the forces and moments in your FBD, and determine the work done by each of those forces and moments. 2. Reaction forces (do not work) 3. If a wheel is rolling on a surface, Then Ff does not work. Ff only does work during sliding/ slipping 4. Look at each force & moment in FBD to determine work done Friction forces do not do any work if you are rolling, Normal force never does any work, internal force never does any work. reaction forces never do any work The work done by the moment, is simply the moment M * the angle the rigid body goes through 5. Total work = sumui , to WED One you have total work, that total work is going to be used in the work Energy principle. Tricks*) 1.) Forces that do no work: F and ds Forces do not do any work, when the distance traveled by that force is 90 degrees to that force. No work. Work is only done by that force, when the distance traveled by that force is in the same direction as that force. 2.) Reaction forces do not work since no displacement Reaction forces do no work, because the reaction forces dont move. The distance of the reaction force is zero. So their is no work 3.) Normal forces never do work Because the normal force, by definition is 90 degrees to the distance traveled. 4.) Friction forces do not work, when rolling occurs Friction forces do not do any work when you have a tire rolling on a surface Only when you have slipping, or sliding does the friction force do any work. When you're rolling, your tire doesent heat up as much compared to when your sliding on the surface. That's why the friction force does not do any work, when you are rolling. 5.) Friction force does work, during sliding or slipping 6.) Internal forces do not work 7.) Reaction forces do not work Always look at the distance traveled by the force, that will tell you if their is a distance, it'll tell you the work done by that force. If that force does not move, their is no work Highlights*) -.) Ff does no work during -.) Ff does work during sliding/ slipping -.) F and D do no work ( N, Reaction forces ) -.) Um = Integral (theta1 to theta2) mdtheta The work done by the moment, is m * the angle theta. The angle theta is measured in radians. -.) Us = -1/2 k(s2^2 - s1^2) ; S2 = L2 - Lo ; S1 = L1 - Lo The work done by the spring, is 1/2k(s2^2-s1^2) -.) Uw = +- mg deltay The work done by mg, is +- the weight mulitplied by the vertical distance The vertical distance of the mass center point g, from state 1 to state 2. Hints*) -) We have done EOM for translation, pure rotation, and general motion. -) If the problem asks for a force, or an acceleration, than it's EOM -) The next topic we're going to discuss is work-energy distance. If your problem has distance in it, think of using the work energy principle (WED). -) The work energy principle for a particle is T1 + sum of all the work done in state 1 and state 2 is = to kinetic energy in state 2 -) Before we can apply the work energy, first we have to figure out how to do the work. The work of a force. -) Rigid body is going to have both forces and moments. You can have a force and a moment on a rigid body. -) For work of a force: we have spring force (Fs), we have weight (mg), we have normal force (N), we have external force (Fα * t ), and we have friction force (Ff). These are all the forces we have seen so far, in the FBD. -) If you want to detrmine the work done by a force, the work of a force is Force * distance. Look at the force, and look at the distance that, that force travels. -) The product of the two is = the work -) The work is only done, along the direction of the force. -) If the force travels 90 degrees to the direction of the force, their is no work If the distance traveled by the force is 90 degrees to the force itself, than no work -) Theirs only work done on the distance, along the direction of the force. -) When you have a rolling problem, their is no work by the friction force. -) Only when you have a slipping or sliding problem, is work done by the friction force -) The work of a moment is Integral mdtheta -) To figure out the work done by a force, look at the distance traveled by that force. -) If you're looking at the work done by the weight, look at the distance traveled by the mass center point g, vertical distance only. Only look at the vertical distance of the weight vector. -) If 'g' moves up, then work is negative. If 'g' moves down, then the work is positive. -) Normal forces do no work, because the normal force is 90 degrees to the distance traveled. -) Friction force does no work, if you are pure rolling -) The work of a moment, is M * the angle The moment * the angle

12.10 Relative Motion

Check List*) 1. Identify two particles 2. Determine the Fixed Coordinate frame 3. Determine position vector of both particles with respect to fixed coordinate frame 4. Calculate Relative motion (All Vectors) Ra with resp. to b, velocity of a with resp. to b, and acceleration of a with resp. to b. Tricks*) 1.) There is always relative motion between two particles that are in motion Their is always going to be relative motion when you have two particles that are moving between each other. 2.) When solving for rA/B + rB, the two B variable's cancel out. making it rA. Also, r A/B = rA - rB Hints*) -) So far what we have seen is rectilinear motion, motion along two axis, the n-t frame, the polar coordinates, and projectile motion. These are all motion of a particle in space. These are all with respect to a fixed coordinate frame -) Except the n-t coordinate frame. the n-t coordinate frame is moving with the particle. -) Dependent motion, is the motion of two particles. -) So far everything that we have done is with respect to a fixed coordinate frame -) ("When you're driving in a car, and you look at the spedometer, and the spedometer says 60 mph. Thats the square root of (vx^2) + (vy^2), the speed. Thats the speed of the car with respect to someone that's fixed to the ground.") -) ("If you're friend has a laser gun and is fixed to the ground, and points it at your car thats going 60 mph. That's the speed that it would record") -) ("Their are two cars that are driving at different speeds, One car is driving 60 mph with respect to a fixed person. Another car is driving 40 mph, the driver looks at the spedometer and sees 40 mph. 40 mph with respect to the same fixed person.") -) Two cars moving at different speeds -) what would the velocity of one car be (Va/b), with respect to another car that's also moving (Vb/a)? That is what relative velocity/ relative acceleration is all about. -) A truck is pulling a crate over a pulley. Their is relative motion between the truck and the crate. -) If someone was standing fixed to the ground, and pointed a lazer gun right at the crate. -) Absolute motion: Motion with respect to a fixed coordinate frame. -) Relative Motion: Motion of one particle, with respect to another particle -) Relative Motion V B/A: Velocity of B with respect to A -) Relative Motion V A/B: Velocity of A with respect to B -) Relative Motion Eq. Va→ V a/b→ + Vb→ -) Relative Motion Eq. Vb→ V b/a→ + Va→ -) Relative velocity vs. absolute velocity. Absolute velocity: fixed coordinate frame Relative velocity: moving coordinate frame -) You have car a and car b, two cars moving along two different paths. You have a fixed observer, the position of a with respect to the fixed observer is ra. The position of b with respect to the fixed observer is rb. The vector that goes from a to b is the position vector of b with respect to a. (Whatever is at the bottom, is with respect to). -) Train a is moving at 30 km/hr, train b is moving in the opposite direction at 60 km/hr. What is the relative velocity? Their are two relative velocities. Va/b or Vb/a. Va, the velocity of train a with respect to someone sitting in train b. Vb, the velocity of train b with respect to someone sitting in train a. (Whatever is at the bottom of that /, is with respect to) How to determine va/b. use the trick, va/b + vb = va and va/b = va - vb What is velocity of a, wrt to fixed coord frame? +30 What is velocity of b, wrt to fixed coord frame? -60 va/b = 30-(-60) = 90 km/hr vb/a + va = vb ; vb/a = vb - va = -60-30 = -90 km/hr -) The Velocity of B relative to A is defined as vB - vA -) Since two dimensional Vector addition forms a triangle, there can be at most _________ unknowns (either magnitudes and/or directions of the vectors)? Two

13.5 Equations of Motion (n-t coordinates)

Check List*) 1. Motion on curved path is n-t Solving problems with n-t coordinates*) - Use n-t coordinates when a particle is moving along a known, curved path. - Establish the n-t coordinate system on the particle. - Draw free-body and kinetic diagrams of the particle. The normal acceleration (an) always acts "inward" (the positive n-direction). The tangential acceleration (at ) may act in either the positive or negative t direction. - Apply the equations of motion in scalar form and solve. - It may be necessary to employ the kinematic relations: at = dv/dt = v dv/ds ; an = v2/rho Hints*) -) Pendulum, If i wanted to detemine the tension in the rope why would i even care to know what the tension is? you want to determine the tension, to figure out if the string is going to hold that barb. because if the string doesent have the strength to hold the barb then it will tear -) If I wanted to determine the tension in the string, how many particles? -) remember FBD, force, tension force, force, FBD, KD force or acceleration , FBD or KD. If i want to find the tension in this, how many particles do I have, that have mass, you have one where would I draw my system If i want to determine tension? I would draw my system here, the weight of the barb (Wb). what would i have if i cut the string? I would have the tension force (T). -) now we are done with the FBD -) Lets figure out the KD, how many accelerations do i have for KD. I have two again. thats mass multiplied by the tangential acceleration (Mat). the other one is mass multiplied by the normal acceleration (Man). -) If i wanted to detemrine the tension? which direction would u start? the normal, because the tension is in the normal drection. Im going to start from normal, sum of all forces in the normal direction (sumFn). (direction of normal dir. is positive dir.) -) Show me the direction of the normal direction, the positive direction. I have (T) that is positive, I have the component of Mg. -) if this is theta, this is theta the component of the weight in the normal axis plus or minus m of the barb, g multiplied by cos theta. that is equal to Man (Normal acceleration = v^2/rho). -) In this problem I dont know what theta is, and I dont know what v is. -) This problem tells me that theta is equal to 30 degrees. So theta is no longer an unknown. -) look for how many equations and how many unknowns you have. -) this says, one equation and two unknowns. What is the next step that you would do? the tangential direction -) sum of all forces in tangential direction (sumFt), tangential direction positive this direction. -) do i have any forces in the tangential direction? Mg sintheta = Mat. (M and M drops out), that gives you the tangential acceleration(at) = to gsintheta. at = gsintheta this gives you the tangential acceleration. -) What is the tangential acceleration? the change in (v) change in the velocity. tangential acceleration is at = dv/dt, meaning that atdt = dv. we have t, and t on both sides. If you integrate zero to t. Im assuming time t is equal to zero, the velocity is zero. This is the velocity at time to. the velcoity at time t is equal to gsintheta multiplied by t is (v(t) = (gsintheta)t) -) Ive gotten the velocity as a function of time -) v is equal to ds/dt. that means that ds is equal to vdt On the left side im integrating over position, on the right side im integrating over time. This is posiiton zero, and this is some position s. what is s? s is Rdtheta (Rdtheta). the s is the positon along the arc, the bob is moving along the arc. the position s is s times rdtheta. -) we're going to say that ds is rdtheta. If i subtitute that into an integral, I have rdtheta zero to theta. that is equal to the integral of zero to time t. gsintheta tdt. Can we take this theta and move it to the other side? no because theta is changing with time. -) The "normal" component of the equation of motion is written as sumFn=man, where sumFn is referred to as the Centripetal force -) The positive n direction of the Normal and Tangential coordinates is Normal to the tangential component, always directed toward the center of curvature, normal to the Bi-normal component

16p1-16p3 - Rotation about Fixed Axis

Check List*) 1. Rotation Axis If it is a rotation problem, try and determine the axis of rotation. 2. Rotation acceleration constant or variable (f(t) or g(theta)) Once you have the axis of rotation, Is the acceleration constant? or changing with time? -) Acceleration constant, use Constant rot. acceleration eq. on equation sheet. -) Acceleration variable, use the variable rot. acceleration eq. on equation sheet. (These equations are analogus, to the equations in the linear world) 3. Vp = omega x rop 4. ap→ = α→ x rop→ - omega^2 rop→ Tricks*) 1.) Identify rotation about an axis 2.) Anolog between pure rotation & translation Their is a direct analog. a 1:1 analog between pure translation and pure rotation. Rotation is nothing but pure translation in angular form Hints*) -) Is the accleeration constant, or is it varying with time? or is it varying with position? -) That tells which set of equations to use, Constant acceleration v = u + at and s = so + ut + 1/2 at^2 -) Or if the acceleration is variable, as a function of time. Then you have a = dv/dt ; v = ds/dt -) If the acceleration is a function of distance, then you have ads = vdv -) Make sure you are very comfortable with this. If you are comfortable. Then pure rotation is going to be no problem -) When you have a rigid body moving through pure translation. moving along one axis -) What can you say about the velocity of A and the velocity of B? They are going to be identical -) If you have a rigid body that is moving in pure translation. Every point on that rigid body has the same velocity. That is the definition of pure translation. -) For pure rotation, the velocities are not going to be the same, but what is the same is the rotational speed. omega at 'a' is going to be the same as omega at 'b' and omega at 'c'. -) Translation world, velocity the same -) Rotational world, Omega the same -) Their's an analog between velocity and acceleration. Velocity being linear motion. Omega or rotational velocity being angular motion. -) In the translation world we have linear acceleration. Acceleration is the first thing I would determine If I had motion translation. -) Where you have acceleration, in the rotation world its rotational acceleration (α). (rad/s^2) Velocity in the linear world is rotational velocity (omega). (rad/s) Translation (S) in the linear world corresponds to theta in the angular world/ rotation world (rad). -) Theirs a direct analog between the translation world and the rotation world. -) The trick for the direction of Omega is the right hand rule. -) If you curl your fingers in the direction of omega, your thumb will show you the positive direction of omega -) If omega were clockwise, your fingers would show the direction of Omega and your thumb shows the axis, or the direction of the vetor. Omega, if you have the x and y. Then z is pointing right at you Then Omega, your finger is pointing in the counterclockwise direction. Omega is positive -) Omega is in the positive direction -) Velocity is always going to be 90 degrees to the distance vector from 'O' to the point of interest. The velocity vector is going to be in the same direction as Omega -) If Omega is counterclockwise, the velocity vector is going to be pointing in the counterclockwise direction. -) If Omega is counterclockwise, what direction is the velocity going to be in? Remember, velocity is in the same direction as Omega. -) Omega cross r (omega x r), is equal to the velocity -) Train yourself to identify problems that have pure rotation. -) Once you have pure rotation, then you know you have one omega and one alpha. Rotational acceleration. -) Their is a direct analog between the translation world and the rotation world -) Remember the velocity is always tangential to the path. -) How many components of the acceleration do I have for p? Two, they are the tangential and normal The tangential is going to be in the same direction as the velocity. -) Omega is velocity, Rotational acceleration is Alpha -) What direction is the normal acceleration going to be? towards the center of rotation. -) normal acceleration is ( an = v^2 / rho ) -) v is omega multiplied by rho ( v = omegarho ) omega multiplied by the radius is v -) Velocity of p. (vp→ = omega→ x orho→) -) Acceleration of p. (ap→ = α→ x op→ - omega^2 op) -) Right hand rule (omega→ x r→ = v→) (x,y,z ; omega, r, v) (Right hand rule) -) Linear Motion = Linear acceleration (a→) Angular Motion = Angular acceleration (α→) -) If you ever get confused by the linear world and the angular world. If you ever get confused by the angular world, look at the linear world. If its a distance in your problem, in the angular world its going to be theta If its a velocity in your problem, in the angular world its going to be angular velocity or rotational velocity -) In the linear world, the first thing you have to do is determine if the acceleration is constant for a function of time or for a function of distance -) The same thing is true for the angular motion, the angular motion is totally analogus to linear motion. -) Variable can either be a function of time or distance f(t). In the angular world their is no distance, its theta f(theta).

15.5 - 15.7 Principle of Angular Impulse & Momentum

Check List*) 1. Rotation problem (Moment involved + Time) If their is a moment involved, its going to be a rotation problem. 2. Axis of Rotation Identify the axis of rotation, and identify the reference points. 3. Point where moment acts 4. Angular momentum along axis of rotation about point where moment acts You have angular momentum because of the moment. The angular momentum determine about which axis the rotation is taking place. and determine the center of rotation. Once you have done that, if their is time involved then you can use AMIT (Angular Impulse Momentum TIme) 5. sumM along axis of rotation about point where moment acts. Tricks*) 1.) Rotation (Particle rotation about an axis) If you see rotation, think of angular motion. Where you have linear momentum, the reason we call it a linear momentum is because we had a force. Angular momentum is going to be because of a moment 2.) Linear (F): m, v, a, f, Linear Momentum (mv), Linear impulse (Integral Fdt) Linear is Force (F). linear motion comes from a force -) This is a trick for you to remember, the difference between the linear world and the rotation world. -Mass (m) (In the rotation world it becomes inertia (I)), -Veloicty (v) (In the rotation world it becomes rotational velocity or angular velocity (omega)) -Acceleration (a) (In the rotation world it becomes angular acceleration or rotational acceleration (α)) -Force (F) (In the rotation world it becomes moment (M)) -Linear momentum (mv) (In the rotation world it becomes angular momentum) -Linear Impulse (Integral Fdt) (In the rotation world it becomes angular Impulse (Integral Mdt)) -) Rotation (M): I, omega, α, Moz, Angular momentum, Angular Impulse (Integral Moz dt) Rotation is moment (M), rotation comes from a moment -)This is a trick for you to remember, the difference between the linear world and the rotation world. 3.) Rotation → Angular Momentum & Angular Impulse If you have a rotation problem think of angular momentum and angular Impulse. Rotation + time ; If you have rotation and time in your problem. Think of AMIT. (Angular Momentum Impulse Time) 4.) What do you need to call angular momentum -) Axis of Rotation (z axis) -) Reference point (center of rotation) For the Axis of rotation, and the Center of rotation (The reference point) If you change the reference point, the moment changes. If you change the reference point, the angular momentum is going to change. -) In our problems, we are only going to be seeing one axis of rotation. Which is the z-axis Because you're working in a plane. The plane has x and y, and the z-axis is coming straight out of the plane. -) you're only going to be seeing rotation out of the z-axis. The first thing to do, is to identify that you have rotation. The second thing, is to identify rotation about which axis. This is a very important trick, to know how to solve the angular momentum Impulse problem. 5.) Always along the 'z' axis It is always going to be the Z-axis, and then identify the center of rotation. 6.) Reference point same on LHS and RHS of Equation 1 The reference point has to be the same on the left and on the right. You cant take the sum of all moments about O, and then take the angular momentum about another point P. That would not work. 7.) Reference point is center of rotation. Make the reference point your center of rotation for Amit problems Hints*) -) Last class we did impact -) Along the x-axis we have COM and COR, BAAB21, A hits B, Along the y-axis we only have COM. The x-axis is 90 degrees to the common tangent. -) When we apply a force on a particle, it either moves along a striaght line or along a curve. Linear motion, Curvilinear motion -) What happens when we apply a moment? -) ("If you were to apply a moment on this car fixed to an axis, what is going to happen to this car? Its going to start rotating. The car will start rotating about that axis") -) The axis at which you are apply the moment (torque). -) What happens when I apply moment to a spindle, that has a bar connected horizontally to its center? It will begin to rotate. -) When you apply a moment, the particle/rigid body starts to rotate. -) Force, results in a linear motion or a curvilinear motion. -) A moment, results in Angular motion Angular motion is the same as rotation, another word for rot. -) What is the linear Impulse? Integral f/t (Integral of force over time) -) If their is a linear impulse, then for angular motion. Their has to be an angular Impulse. -) What do u think the angular Impulse would be? Integral m/t (The integral of moment over time) -) Linear momentum? Linear momentum is mass * velocity (mv) -) What is Angular momentum? -) A force is going to result, in a curvilinear motion. Or a linear motion. -) Moment is going to result in something rotating. -) When you have a force. when you have linear motion or curvilinear motion. You have linear momentum and linear Impulse. -) When you have angular motion, rotation. Then the linear momentum becomes angular momentum. -) Linear Impulse becomes Angular Impulse, why? Because rotation is nothing but angular motion. -) MIT and angular MIT (aMIT). -) Angular Momentum -) Angular Impulse is, angular means moment. Angular motion only comes from a moment -) Linear motion comes from a force. Linear Impulse is the integral of FDT ( Integral FDT ) -) Angular Impulse, since it comes from a moment is the integral of Mdt (Integral Mdt). -) Linear momentum is mass * velocity (L=mv) -) If you have a particle? and you have a force applied to that particle. Then the moment of that force -) When your taking the moment of a force, you have to ask yourself, moment about which point. The moment about A ; The moment about origin ; The moment about b or p or q. -) For the moment, you need one more piece of information about which point? The moment about O, Is the position vector from O to the point p. The position vector from point O to point p x F. (x=crossprod.) -) What would the moment with respect to q be? (Mq→ = rQp→). The vector that goes from q to p. cross (x) the moment of this force. (Mq→ = rQp→ x F→). -) That's the moment, of this force with respect to q. -) Lets look at linear momentum, recall linear momentum has nothing to do with moment. -) Momentum has nothing to do with moment. Linear momentum, of this particle P. is mass * velocity. (P = mv) -) Velocity is always tangential to the path, so the linear momentum is gonna be in the same direction as velocity. -) The velocity is always tangential to the path. m*v is the linear momentum. -) Therefore, the linear momentum is also always tangential to the path. -) If the linear momentum is always tangential to the path? Then what is the moment of the linear momentum. -) Its a vector, Linear momentum is a vector. Just like force is a vector. -) you determine the moment of that force. Determine the moment of the linear momentum -) The moment of the linear momentum, is the angular momentum. -) The same equation as the moment here, The angular momentum of our O (Ho), Is rop→ x L→ ; (Ho→ = rop→ x L→) -) That is the angular momentum -) What would the angular momentum (Ho→ = rop→ x L→) of this particle p be with respect to this point q? (HQ→ = rQp→ x L→) -) you have linear momentum (L→ = mv→). Its a vector If you think of that vector as a force, its not a force. If you think of it as a force, then you can take the moment of that linear momentum with respect to any point. By taking the x (cross product) of the vector from the point of reference to the point of interest, and x (cross product) it with the linear momentum vector. -) Remember for M (Moment), moment is a vector. For M you need two pieces of information? about which reference point, The moment about which reference point, and the moment about which axis. -) What is linear momentum? ( L→ = mv→ ) what is linear Impulse? ( I→ = Integral Fdt→ ) -) What is angular momentum? ( H→ = r→ x mv→ ) ( r cross linear momentum) What is angular Impulse? ( Integral m→ dt ) ( Integral of m over time ) -) What is angular momentum? Angular momentum is the moment of the linear momentum. -) What do you need to calculate Angular Momentum? The Axis of rotation, and the Center of rotation (The reference point) If you change the reference point, the moment changes. If you change the reference point, the angular momentum is going to change. -) In our problems, we are only going to be seeing one axis of rotation. Which is the z-axis Because you're working in a plane. The plane has x and y, and the z-axis is coming straight out of the plane. you're only going to be seeing rotation out of the z-axis. -) This is a very important trick, to know how to solve the angular momentum Impulse problem. -) For the reference point, look for the center of rotation of that particle. -) This is the challanging part of Angular Momentum Impulse. -) The first thing to do, is to identify that you have rotation. The second thing, is to identify rotation about which axis. -) It is always going to be the Z-axis, then identify the center of rotation -) look at your P&F problems, dont solve. Just identify the axis of rotation and the center of rotation. -) Units for Angular Momentum, [m kg m/s] ; [ kg m^2/s ] -) Principle of Angular Momentum -) Remember, moment reults in rotation Force results in , linear motion or curvilinear motion. Force and velocity are connected. Moment and angular momentum are connected. -) Force is connected to linear momentum, force is linear motion. -) Moment is connected to angular motion (rotation). Therefore moment is connected to angular momentum. -) (MIT) Angular momentum impulse principle. -) Linear momentum principle? Mv1 + Integral F dt = mv2 -) The analog of this equation is, instead of linear momentum. You have angular momentum + sum of all moments integrated over time is equal to the angular momentum of state 2. ( Mv1 + Integral F dt = mv2 ) -) The left side is equal to the right side -) Remember, the moment you need with respect to what axis about which point. -) Theirs only one axis of rotation ,the z-axis -) The point of reference has to be the same on the left side and the right side. -) Angular momentum is a moment, you have angular momentum along the z axis with reference point O. Sum of all the moments about the z axis with reference points O. Sum of all the moments about the z axis, in state 2. Reference point O -) Another trick to remember, always along the z-axis. The rotation is only along one axis. And the reference point on the left hand side O, must be the same as the reference point on the right hand side. -) This O, could be p, q, or r, a different point. You have to take the sum of all the moments about that. The same point -) If this is P, this becomes P, and so does this -) The reference point has to be the same on the left and on the right. -) you cant take the sum of all moments about O, and then take the angular momentum about another point P. That would not work. -) Make the reference point for Amit problems, make the reference point your center of rotation.

Exam 1 Review

Check List*) 1. Understand the problem 2. Identify the type of problem 3. Fix your coordinate frame 4. Break up the problem into sub-parts 5. Use appropriate equations 6. Remember units in final answer, not intermediate calculations Hints*) -) So far we've looked at the motion of single particle's, and we've looked at the motion of two particles. -) The moment you have two particles moving, you also have relative motion between the two particles. -) If you only have one particle moving, its absolute motion with respect to a fixed coordinate frame. -) ra→, rb→, the position vector from the origin to particle a, to particle b. -) Position of b with respect to a is a vector (rB/A→) . starts from the bottom and ends at the top. starts at A and ends at B. -) The same thing for rA/B →, the Position of a with respect to b, starts at B and ends at A. relative acceleration -) All of the material we've learned so far can be summarized by a = ds/dt ; v = ds/dt -) Their are various forms of these two equations. If the acceleration is constant. these equations become the constant acceleration equations. -) These two equations are the generic equations, a = ds/dt ; v = ds/dt -) For a single particle, you have the bubbles n-t (Pos, Vel, Accel), Rectangular (Pos, Vel, Accel), and r-theta (Pos, Vel) It all starts with the coordinate frame. You have to fix your coordinate frame based on the type of problem -) If you have a curved path, and rho (rho) is given. The radius of curviture is given. Then most likely you will use the n-t coordinate frame. ( n, t, b coordinate Eq.) -) If you have a curved path, and r, and theta, and theta dot, or r dot is given. Then most likely you're going to be using the polar coordinates, r-theta ( r, theta, z coordinate Eq.) -) If its x and y, if its rectangular coordinates. Then x and y represents the rectangular position along the x and along the y (x, y, z coordinate Eq.) -) For (SIngle particle, Two particle, Coordinate frames, rectangular, n-t, and r-theta) ; a = ds/dt and v = ds/dt apply. -) The velocity vector, regardless of coordinate frame. Is always going to be tangent to the path, regardless of coordinate frame -) The acceleration vector, regardless of coordinate frame always has two components. The tangential component and the normal component. regardless of coordinate frame. Scope for Exam 1 Rectilinear Motion Erratic Motion Motion - rectangular coord Projectile motion n-t Coord Frame r-theta Coord Frame Dependent Motion Relative Motion

16.7 Relative Motion Analysis: Acceleration

Check List*) 1. Verify General motion all this relative stuff, does not apply if you have linear translation or curvilinear translation. or pure rotation. 2. Relative motion analysis - velocity to determine omegaab First verify which link is going through general motion 3. Relative motion analysis - acceleration to determine αAB, αa, αb accleration is a two step process, first you need to do the velocity. From the velocity analysis you get the acceleration analysis. For acceleration you need the velocity analysis to determine omega. Tricks*) 1.) Velocity analysis, 3 velocity quantities needed For the velocity analysis, remember you need 3 quantities, 3 velocity quantities. You need magnitude, direction, and either magnitude or direction 2.) Acceleration analysis, 3 acceleration quantities needed You need 3 acceleration quantities to do the acceleration analysis 3.) For acceleration, 2 step process - Velocity analysis - Acceleration analysis Hints*) -) So far we have done pure translation, their are two forms of translation -) Translation along a straight line, and curvilinear translation -) We did pure rotation, the trick their is the velocity is always 90 degrees to that link -) After pure rotation, we combined both translation and rotation to get general motion -) General motion is equal to relative motion -) Their are two ways of solving a general motion problem The first method is relative velocity method two is instananeous center of rotation. -) Today were going to discuss the acceleration of a point thats going through general motion. -) So far we have the position, velocity. All we need is the acceleration. -) How do I determine the acceleration of a point. We know the position, we know the velocity. How to determine the velocity of a point on a rigid body going through general motion. -) Mass * acceleration is = sum of all the forces. -) How many components of acceleration are their? Two, normal acceleration and tangential acceleration. -) The normal acceleration points towards the center of rotation. If you have pure rotation, the center of curviture is the point about which the arm is rotating. The normal acceleration is gonna be pointing towards the center of curviture. The tangential acceleration is 90 degrees to that link. -) General Motion acceleration analysis. aB = aA + aB/A = aA + αAB x rB/A - omegaab^2 rB/A What you have on the left must match what u have on the right

14.5 - 14.6 Conservative Forces, Potential and Energy

Check List*) 1. Verify no friction & no external force Verify you have no friction and no external forces acting on the particle. except for mg and spring force 2. Define state 1 and state 2 define your states, state 1 and state 2 3. Define datum for Vg define the datum for the gravitational potential energy. 4. sumT1 + sumV1 = sumT2 + sumV2 Sum of all kinetic energies in state 1 + Sum of all the potential energies in state 1 = Sum of all the kinetic energies in state 2 + Sum of all the potential energies in state 2 (sumT1 + sumV1 = sumT2 + sumV2) If you have multiple particles. Than each of those particles, If they have a velocity and a mass. It has Kinetic Energy, you would take the sum total of all kinetic energies. And the sum total of all potential energies. Remember 1 Datum, datum is fixed Tricks*) 1.) If your FBD only has mg, and/or a spring force (Fs). You can use Conservation of Energy (COE) Hints*) -) Last class we looked at WED (Work Energy Distance) -) If you see a problem that has distance, most likely you would start with WED. -) The mother of all equations in dynamics is Newtons second law. (sumF→ = ma→) -) From Newton's second law, you can derive the work-energy principle. T1 + sumu1→2 = T2 (WED) From the work energy principle, if you have no friction in your problem, and if you have no external forces -) If your FBD only has mg, and/or a spring force (Fs). You can use Conservation of Energy (COE) -) Conservation of Energy becomes the kinetic energy in state 1 + the potential energy of state one is = the kinetic energy of state 2, plus the potential energy in state two. -) If you look at the pendulum, this is state 1 and state 2. If you look at the FBD of the pendulum, you only have Mg force. Yes you also have the tension force, but thats if you break the rope with your system boundary. But if your looking at this entire system. This entire pendulum as one system. You're boundary is basically this. If that is your system, then the tension force becomes an internal force. and that doesent show on the FBD -) Mg is the only force you have, Remember, if you have Mg and/or the spring force only. Then COE applies. -) Between state 1 and state 2, the kinetic energy + the potential energy is equal to kinetic energy + potential energy in state 2. -) If you're free body diagram only has weight, and/or spring force. Think of using, COE -) The potential energy has two parts to it. The potential energy has the gravitational potential energy. And the elastic potential energy. -) The gravitational potential energy is positive, if the particle is above the datum. The gravitational potential energy is negative, if the particle is below the datum. -)Whenever you're doing the COE, and you're defining the Potential energy (PE), make sure to define the datum. -) The datum, you can define wherever it is that you want. -) We know the kinetic energy in state 1 is T1 = 1/2 mv1^2 -) For potential energy, the moment you you talk about potential energy. You have to define a datum. -) You can define the datum wherever it is you want to, as long as it is fixed to the ground. -) If the mass is below the datum, you're datum will be negative -) Potential energy will be positive if the particle is above the datum, potential energy will be negative if the particle is below the datum -) The potential energy does not depend on the path, that a particle takes. wether it goes on one path, or the other path. Potential Energy is mass * g * h (The height) -) Elastic potential energy is 1/2 K * stretch^2. -) Stretch is the new length - the original free length. S1 = L1 - Lo -) Stretch in state 2, is the length of the spring in state 2 minus the original. S2 = L2 - Lo -) Remember the elastic potential energy is 1/2 ks^2 -) Mg = Gravitational force -) The Potential energy of a spring is Always positive -) When the Potential energy of a conservative system increases, the kinetic energy Always decreases

12.6 Projectile Motion

Check List*) 1.) Define X & Y axis 2.) ax=0 & ay= +- g 3.) Seperate Motion along X and Y Tricks*) 1.) X axis is → direction of forward motion Y axis is → direction of Gravity Identify the two Axis. The Y-axis by definition for projectile motion is Up and Down. The X-axis is foreward motion 2.) Motion in X & Y are dependent on each other because y(x). In projectile motion, the motion along the x and y are dependent on each other. 3.) There are two unknowns, because there are two equations (Only two of these three equations can be used) 4.) Time common to motion along X & Y Hints*) -) Projectile motion is a special case of Rectangular Motion -) When you see a problem you should be able to identify if its Rectilinear motion? Rectangular motion? or Projectile motion? (Train your mind to do this), Only then can you apply the proper tool from your equation sheet -) How do you identify projectile motion? for Projectile motion their are no forces acting on the particle. their is no thrust acting on the particle. ("If you have an airplane taking off, that is not projectile motion because you have the engines providing thrust. The moment the engine and thrusters are shut off, the airplane goes into projectile motion") -) Their are only two Axis for Projectile motion, because we're looking at it in a plane. (X axis, Y axis) -) ax = 0 ; ay = +-g The acceleration along the X axis for projectile motion is always zero (ax = 0). The acceleration along the Y axis is either positive g or negative g (ay = +-g), depending on the orientation of your Y axis. -) g is always pointing down regardless of the problem. If your Y axis is pointing up then ay is negative g. If your Y axis is pointing down then ay is positive g. (Accel, Vel, Pos. are all vectors) -) For projectile motion, the projectile is constrained to move on this equation. Y = (Xtantheta) - ((gx^2)/(2vo^2)) * (1+tantheta^2) At any point if you know what X is and you know what the angle theta (The initial angle) is. Then you can solve for the Y -) When you look at projectile motion, notice if the acceleration is constant? Or if the acceleration is variable? -) The acceleration along the X-axis is zero, meaning the velocity is constant -) The velocity vector is always tangential to the path, meaning the velocity vector has an X-component and a Y-component. So the speed of the particle is sqrt(vx^2 + vy^2). When you have projectile motion, every point on that porabola has an X-component of velocity, Y-component of velocity. -) How to determine the X and Y component? for the vertical motion the acceleration is ay = -g (Use constant a = ac equations) [Vy = Voy - gt] (Its '-' because the accel. is negative) (Voy = initial Vel. of the particle at time t=0) [y = yo + (Voy)t - 1/2 gt^2] [Vy^2 = Voy^2 - 2 g (y-yo)] Out of these three equations, how many are completely independent of each other? Only two are completely independent. You can only use two equations out of three -) For projectile motion, constant acceleration (acceleration zero along the X axis), and acceleration that is -g along the Y axis. You only have two equations that you can work with. If you only have two equations, how many unknowns can you solve? Two eq : Two unkowns ; Three Eq : Three unknowns -) The downward acceleration of an object in free-flight motion is 9.81 m/s2 -) The horizontal component of velocity remains _____ during a free-flight motion? Constant

12.1 - 12.2 Intro Rectilinear Kinematics: Continious Motion

Check List*) 1.) Is it a Particle or RIgid Body 2.) a=c or a=f(t) → (Is acceleration constant?, varying with time or position?) 3.) If a=f(t) → a = dv/dt (If acceleration is varying with time) If a=f(s) → ads = vdv (If acceleration is varying with Position) If a=c → (Constant a = ac) 4.) you only have one Coordinate Axis and one Origin 5.) Units Tricks*)= 1.) If you see particle or rigid body moving. Is Acceleration constant or variable? Is acceleration changing with time or position? 2.) If your problem has time in it use acceleration variable equations 3.) If your problem has position in it use ads = vdv (Equations are for non-constant or variable accel.) Hints*) Ch.13 -)Force/ Acceleration -) Variable Acceleration is above constant acceleration -) You can derive the constant acceleration equations, from the variable acceleration equations. -) In Dynamics, a particle is assumed to have Only a mass -) The average speed is defined as sT/∆t Ch.14 -) Non Impact distance -) No impact, no contact, no collision -) If you have no impact, and if you have distance Remember WED. -) If you do use WED, make sure you define the states. state 1 & state 2. What is state 1 & what is state 2 -) What is the challanging part about work-energy? Energy is clear, that is (1/2m v1^2 + 1/2m v2^2). -) Then comes the work by all the forces in our FBD.

19.4 Rigid Body Impact

Hints) -) Rigid body Impact so far we've done impact for particles -) The line of Impact LOI 90 degrees to the plane of contact -) All of this applies to rigid body Impact -) Impact is when two rigid bodies come into contact with each other -) The moment they come into contact with each other, You have Impact equations If they're seperated from each other, its work-energy. Sum of all the forces It's anything else but Impact. -) Impact takes place over a short duration of time. -) The plane of contact? The plane of contact is a plane that is tangent to both particles. -) If you look at the plane of contact for two rigid bodies coming together, that plane of contact is coming in and out of the paper. The line of impact is 90 degrees to the plane of contact. -) You have central impact, when the velocities are along the line of impact, before and after -) If their is an angle between the velocities either before or after, To the line of impact Then you have Oblique impact Regardless of wether you have central impact or oblique impact. -) For Impact problems remember, along the line of impact you have Conservation of momentum (COM) -) Their are two types of momentum, linear momentum and angular momentum. -) The conservation of linear momentum and you also have another equation called the coefficent of restitution. -) In BAAB21, what hits what? A hits B, in BAAB21 -) Coefficent of restitution or conservation of momentum along the line of Impact. -) Normal to the line of impact, 90 degrees to the line of impact. The velocities before and after impact are the same. -) The only difference between a particle and a rigid body Particle has no dimensions, It doesent mean that the particle has to be a small object. It can be an 18 wheeler, car, bike As long as it has only translation. Particle has only translation. -) Rigid body has dimensions, It has length and width. The radius of the spehere. Rigid body can translate, it can rotate -) The moment a rigid body moves it has energy It has kinetic energy -) If it has kinetic energy, that means that It has a velocity. If it has velocity, then linear momentum. If it rotates, then angular momentum Rigid body also has moment of inertia. -) The only thing a particle doesent have, is mass moment of inertia. Mass moment of inertia doesent exist for a particle.

17.1 Mass Moment of Intertia

Tricks*) -) Force→ translation Moment→ rotation Hints*) -) What is the mass moment of intertia Mass is resistence to translation. The heavier mass you have, the higher is its resistence to moving. -) Moment of intertia, is resistence to rotation. (" You have a slender cylinder, you are going to rotate that cylinder by applying a moment") -) If you had a larger cylinder relative to your slender cylinder. Where would you apply less of a moment? It would be the slender. -) The slender the rigid body is, the smaller is the moment of inertia. -) You need smaller torque, or smaller moment. For a smalller moment of inertia. -) If the object is larger, the moment of inertia is higher. You need more force, more torque, more moment -) The slender cylinder has less mass, if the mass is closer to the center of rotation. The axis of rotation -) For slender cylinders, the mass is closer to the axis of rotation. For huge cylinders, the masses are furthur away from the axis of rotation. For the huge cylinder, because the masses are furthur away, the moment of inertia is higher. -) Higher moment of inertia, means you need higher moment, higher torque to rotate. -) If you see a moment on a rigid body, think of rotation. When you think of rotation, think of moment of inertia. -) Moment of inertia is the resistence to rotation. The higher the moment of inertia. The larger the torque required to rotate body -) The sumF→ = ma→ (Linear motion), the equivalent equation to this except in rotation is sumMg→ = IGα→ (rotation) -) The moment is a little different compared to the force. In the force, all you have to know is what direction were applying, and what is the magnitude of that force. In moment, when your taking the moment of the rigid body. You have to ask yourself what axis? and with respect to what? -) sumMg→ = IGα→ (rotation). I is about one axis, G is 2 reference point -) correction factor is mass multiplied by the square of the distance -) Mass moment of inertia using the radius of gyration The moment of intertia is the integral of the distance of mass from the axis of rotation. That distance^2 * mass ( IG^2 = mKG^2 ), -) If the radius of gyration is given to you by another point O, then ( IO^2 = mKO^2 ) -) In a problem, you can either be given an equation about the moment of inertia about the z axis. with respect to the mass center point G Or the problem is going to give you the radius of gyration.

15.3 Conservation of Linear Momentum

Tricks*) 1.) Contact is constant of Linear momentum and MIT 2.) When deltat is not given (Conservation of Momentum) If the deltat, time at which the force is acting on the particle. If that deltat is not given to you COM (Conservation of Momentum), as long as you look at both particles together. When deltat is not given, think of COM 3.) When deltat is given (MIT) Hints*) -) Last class we talked about Momentum and Impulse -) Momentum is the mass * velocity (mv→) -) Impulse is the integral of the force (Integral Fdt) -) You would use MIT, the T stands for "Time". The MI is "Momentum Impulse". MIT = (Momentum Impulse Time) -) If it's a distance problem WED, its a time problem MIT -) From WED, you can simplify it to Conservation of Energy (COE). You can use COE when their are no external forces and no friction. -) If you have Momentum Impulse, then logically what would you think their is? not Conservation of energy, but Conservation of Momentum -) Remember anytime you see an Impact problem, their is Conservation of Momentum. -) Conservation of Momentum is when you have deltat -) If neither time or distance is given to you, and you have an Impact problem. think of Conservation of Momentum -) WED becomes COE Remember work energy, and momentum impulse. Are the higher level equations. higher than that is sumF = ma. From sumF = ma, comes WED, and MIT. From WED comes COE, and from MIT comes COM (Conservation of Momentum). -) If you have any kind of impact, and the time deltat is not given to you. Then think of COM (Conservation of Momentum) -) What does COM mean? The momentum of my system before impact is = to the momentum of my system after impact. -) If I have two particles that are going to come into contact with each other, right before impact. The momentum of this particle + The momentum of this particle is = the momentum of the particle right after impact. -) Right before impact, the total momentum is = to. Right after Impact. -) Remember, Momentum Impulse is only valid for a short duration of time. Whatever time that the force is acting on, thats when momentum impulse is valid. -) Conservation of momentum, remember 'Impact'. We need impact in order to apply the COM (Conservation of Momentum) -) Conservation of Momentum looks at the entire system. Not just one system. If you only have one system, if you only have one particle, then their is no conservation of momentum their. -) Conservation of Momentum only takes place when you have two or three particles. And you're looking at it as one system. -) Conservation of linear momentum. Look at the entire system, look at both particles. -) Anytime you have impact its COM (Conservation of Momentum) -) Anytime you have impact, you have COM right before and right after. Only for that deltat.

16.5 Relative Motion Analysis : Velocity

Tricks*) 1.) Think vectors for relative motion analysis (velocity & acceleration) General motion = relative velocity. Relative motion equation: VA = VB + VA/B = VB + omegaAB x r A/B 2.) You should be able to identify pure rotation motion & pure translation (Curvilinear motion) This only applies to general motion, translation has no relative velocity for two points. Same think for pure rotation, their is no relative velocity for the two points 3.) Relative motion analysis is only for general motion problems. General motion = Relative motion 4.) How to identify general motion? If two points on the same rigid link have velocities. The moment two points on the same rigid link have different velocities. You have general motion. 5.) Two coordinate frames, (x,y), and the (x',y'). The x prime is parallel to the x. The y prime is parallel to the y. -)It is a Fixed coordinate system (Absolute coordinate system) -)Moving coordinate frame (x',y') attached to point You move the x',y' to the point where you're taking with respect to. If you're taking the vector A with respect to B. Then you're going to be moving this fixed coordinate frame (x,y) to B. Because you want the coordinates of A with respect to B. Trick for Relative motion analysis (velocity) (General Mot.)* 1.) Fix coordinate frame That is the first step in any dynamics problem 2.) Identify rigid body in general motion 3.) Identify velocity vector of 2 points on rigid body in general motion. -) Velocity of one point (MAG & DIR) -) Velocity of 2nd point (Either MAG or DIR) You need 3 quantities to solve the general motion of any problem. You need three pieces of information Velocity is a vector, it has magnitude and direction. Magnitude is one quantity and direction is another quantity. 4.) 2 equations and 2 unknowns (omega, and velocity MAG/DIR) General motion, relative velocity. In order to solve a relative velocity problem you need three pieces of information. Magnitude is one piece of information, and direction is another piece of information. We need three of those for general motion Hints*) -) Motion of a rigid body -) You have three different types of motion only -) Pure rotation, Pure translation, and translation & rotation -) For translation, we have two different types. rectiliner translation, or translation along a curve. -) When you have translation along a curve, that rigid object doesent change angles. It stays parallel. -) Rotation, you have one axis. It rotates about that axis -) What happens when you combine translation and rotation? you get General Motion -) For pure translation, all points in that rigid body have the same velocity. -) For rotation, all points on that rigid body. Have the same rotational speed. Theirs only one rotational speed -) General motion, you have one rotational speed, and mulitple speeds. They are all different. Every point on that rigid body has a different velocity. -) General motion means every point has a different velocity -) When the angle changes its general motion -) When looking at a rigid body try and determine if its Pure translation, pure rotation, or general motion. -) Practice what kind of motion every link in a mechanical system goes through. -) If you have general motion, you have relative velocity -) If problem asks for force or acceleration. that's = FBD or KD -) If you see general motion, immediately think of relative motion. -) General motion is equal to Relative motion -) Relative motion equation VA = VB + VA/B = VB + omegaAB x r A/B Use trick learned in pervious chapters, if you have a/b on the left. you must have a/b on the right -) Whats r A/B, it starts at the bottom and ends at A.

12.4 - 12.5 Motion : Rectangular Coordinates

Tricks*) 1.) Time is common to motion along x, and y axis Time is the only common variable to both ("You only have one stop watch for X-axis and Y-axis, r axis, theta axis ") 2.) Motion along the X is independent of the motion along the Y (In the general Case) ("If you see a rocket moving in space, its X and Y are independent. Because it can move anywhere") If it were constrained to move along a cirlce then the X and Y are connected. If you give me an x, I would be able to figure out the Y (vice-versa). 3.) Velocity vector is always tangential to the path The vector itself is tangential to the path. The speed doesent have direction, but the vector does 4.) Acceleration vector is not tangential to the path of the particle. ("When your driving down a straight road, theirs only one acceleration. But the moment you take a curve, you have two accelerations") 5.) a→ (an: Normal to path ; at: Tangential to path) Only when you have striaght line motion do you have one acceleration component, otherwise you will always have two acceleration components 6.) Only one coordinate frame 7.) Velocity is always tangential to the path The acceleration is not going to be tangential unless you have rectilinear motion. The moment you have any motion that isnt rectilinear, you're going to have two motions (Normal accel. , Tangential accel.) Hints*) -) When you see motion ask yourself, is acceleration constant? or is accelertion varying with time (Same is true for mation along 2 axis). Is the motion along the x-axis constant acceleration or variable acceleration. Is the motion along the Y-axis constant acceleration or variable acceleration. The same question applies to motion along x-y, r-theta, or n-t. -) When you say the temperature is 20 degrees, you dont say it's along the x axis or the y axis. If you say the force f is 20 N, thats not enough. You have to indicate 20 N along x axis or y axis. -) If you see variable speed. We're talking about the square root of the sum of the squares. Sqrt( Vx^2 + Vy^2) -) If we talk about Vector, we're talking about multiple components -) For motion in a circle, the velocity component will always be tangential to the path -) If the tangential acceleration is zero, the velocity is constant -) When the tangential and normal acceleration are zero, you have zero acceleration. -) When you have tangential acceleration zero, the tangential velocity is not changing, its the same -) In curvilinear motion, the direction of the instantaneous velocity is always Tangent to the path -) In curvilinear motion, the direction of the instantaneous acceleration is always Tangent to the hodograph

12.3 Erratic Motion (Study this one alot more, very weak)

Tricks*) 1.) s-t (Use ds/dt) → v-t 2.) v-t (Use dv/dt) → a-t 3.) a-t (Use dv/dt) → v-t 4.) a-s (Use dv/dt) → v-s 5.) v-s (Use ads = vdv) → a-s Hints*) -) Slope is the same thing as derivative. -) When you see dv/dt and dv/ds it's a slope. If you see an integral, its the area under the graph -) The slope of a v-t graph at any instant represents instantaneous Acceleration -) Displacement of a particle in a given time interval equals the area under the (C) graph during that time. v-t P.S) your weakness is the coordinate frame, Integrating and deriving, and bracket mathematics