Exam 3 Problems

How to obtain a momentum vector from a time-dependent position vector: r = #xi+ #x^2j

1. Derive the equation to get velocity equation 2. To find equation for p, multiply the mass in front of both I and j values of velocity equation

A particle is located at xyz coordinates (2.00 m, 3.00 m, 4.00 m). A force given by F→=(5.0 N)i^+(−1.00 N)k^ acts on the particle. (Note that the y component is zero.) We want the torque on the particle about the point with coordinates (−1.00 m, −2.00 m, 5.00 m).

A force F→ can produce a torque τ→ on a particle according to τ→=r→×F→, where r→ is the position vector that points from a specified point to the particle. The torque is said to act about that specified point. The position vector r→ can be determined by subtracting the coordinates of the specified point from the coordinates of the particle, axis by axis. First: Calculate position vector for each axis by subtracting final from initial. For example: rx = (2.00m) - (-1.00)m. Last, recall τ→=r→×F→, use cross product as shown in picture.

Figure shows a uniform disk with mass M = 2.5 kg and radius R = 20 cm, mounted on a fixed horizontal axle. A block with mass m= 1.2 kg hangs form a massless cord that is wrapped around the rim of the disk. In the limit of m→0, what numerical limits do a, T, and α approach?

All zero.

The figure is a plot of the (single) torque τz acting on a particle versus time t. We want the change Δℓz in the particle's angular momentum from t0 = 0 to t1 = 5.0 s.

Area under the curve

Torque on a particle due to force: In this figure, three forces, each of magnitude 2.0 N, act on a particle, which is at point A that is located by position vector r→, where r = 3.0 m and θ = 30°.We want the torque about the origin O due to each force.

Because the three forces do not lie in a plane, we must use cross products to find the torques, with torque magnitudes given by τ = rF sin ϕ and torque directions given by the right-hand rule. Move tail of F vector to tail of R vector. Measure angle/R hand rule that way.

Chain reaction of elastic collisions: Body 1 approaches a line of two stationary bodies with velocity v1i = 10. Body 1 collides w body 2, collides w body 3, mass m3 = 6. V3f is given. Assume elastic collision with no net external force. What is the value of m2?

Can use elastic collision equations (v1f =... v2f =...), Rewrite equations for box 2 and 3. Once you find m2 and v2, you can plug back into original equations. **be careful w notation body 2. V2f just after first collision is same vas v2i from before second collision.

How would you find angular acceleration given theta or w equation?

Derive a twice or w once.

This first figure shows a freeze-frame of a 0.500 kg particle moving along a straight line with a position vector given byr→=(−2.00t2−t)i^+5.00j^,with r→ in meters and t in seconds, starting at t = 0. The position vector points from the origin to the particle. In unit-vector notation, we want to find expressions for the angular momentum ℓ→ of the particle and the torque τ→ acting on the particle, both with respect to (or about) the origin. We also want to justify their algebraic signs in terms of the particle's motion.

Derive position vector to get velocity vector. Take cross product of r and v using cross product rule (can really only evaluate last term) Multiply by mass to get L Torque is the derivative fo thee L. Evaluate at a few different times to see what direction it travels in using the R hand rule.

Torques due to forces on a rod: What is the straight-line distance r1 for force 1 with magnitude F1= 1.0 N? What is the angle ϕ1 between an extension of that straight line and the direction of the force? What is torque due to force 1?

Distance from center to point of force. Angle is 0. Torque is zero: t1 = r1F1sin(θ)1 = 0.40(1.0N)sin0 = 0. Since sin0 is zero. **Measure angle from base of the line to where the force extends.**

A child's top is spun with angular acceleration a = 5t^3-4t with t in seconds and in radians per second-squared. At t=0, the top has angular velocity 5 rad/s, and a reference line on it is at angular position = 2 rad. Obtain expression for the angular position theta(t) of the top.

Find theta(t) by integrating w(t) with respect to time.

This figure shows a 2.0 kg toy race car before and after a turn on a track. Its speed is 0.50 m/s before the turn and 0.40 m/s after the turn. In unit-vector notation, what is the change ΔP→ in the linear momentum of the car due to the turn?

Find x and y vector values by looking at direction car travels at each velocity (if the car is going straight down, x will be zero, etc.). Use change in momentum equation (subtract x values and y values) to get change in momentum.

Figure shows a uniform disk with mass M = 2.5 kg and radius R = 20 cm, mounted on a fixed horizontal axle. A block with mass m= 1.2 kg hangs form a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block.

Fneet = ma, tnet = Ia. Accelerations of wheel and block ar equal since they're tied together. Combine equations T - mg = ma and T = -1/2Ma. Solve for a.

The figure gives the magnitude p of the linear momentum versus time t for a particle moving along an axis as a force directed along the axis acts on the particle. Rank regions according to MAGNITUDE of the force.

Fnet = dp/dp = SLOPE of each region. Steeper slope will be greatest, negative slope that is less steep will be second and slowing down, zero slopes will be last.

For four situations, here is the angular position θ(t) of a rotating body, in radians and seconds. For which can we apply the equations for constant acceleration?

Integrate 3 times. If final answer is just one number, then it is constant.

At time t = 0, a force F→=−2.00ti^ (in newtons and seconds) begins to act on a particle traveling along an x axis with initial linear momentum p→=(5.00 kg⋅m/s)i^. What is the particle's momentum at t = 3.00 s?

Integrate the force with respect to time: In this case, becomes -x^2 + C. Sub in initial and final values.

In the figure, a disk, a hoop, and a solid sphere are made to spin about fixed central axes (like a top) by means of strings wrapped around them, with the strings producing the same constant tangential force on all three objects. Rank the objects according to the magnitude of their angular momentum about their central axis at a certain time t. From Module 10.5, the rotational inertias of the objects are the following: Idisk=0.5⁢ MR2Ihoop= MR2Isphere=0.4⁢ MR2. Rank the objects according to their angular speed about their central axes at a certain time t.

L = FRt, all objects have same angular momentum at time t. Second: Smallest I is biggest angular momentum.

2D explosion: Firecracker explodes inside a coconut, blows it into three pieces. Piece C has mass 0.30M, final speed vac = 5.0m/s. What is speed of piece B with mass 0.20M? Piece A?

Linear momentum is conserved; closed system + internal force (explosion). Superimpose xy coordinates, make one vector equal the axis, changes angles as necessary,. Piy = Pfy **Linear momentum is NOT conserved along x axis, gravitational force is external force acting on system along x. Conserved along y because no external forces acting here.

Conservation of momentum, ballistic pendulum: Bullet hits block of wood hanging from cord. They come to rest on collision the swing upward, COM rising h=6.3 cm before momentarily coming to rest at end of arc. What is speed of bullet just before collision?

ME bottom = ME top. Initial: KE Final: Gravitational potential energy No net external force on the bullet-block system; upward tension force balances downward gravitational force. Not conserved during rise of system. ME not conserved on collision due to sound and friction. Tension force is perpendicular to path so cannot alter ME. ME is conserved during rise because only force doing work on sx is gravitational force (conservative force)

In the figure, an initially stationary block of mass m = 3.00 kg begins to descend as a connecting cord unwraps from a pulley. The pulley, which is mounted on a horizontal frictionless axle, is a disk (assumed uniform) of radius R = 0.200 m and mass mp = 8.00 kg. We want the speed v of the block and the angular speed ω of the pulley when the block has fallen 0.150 m.

No friction, so there is conservation of mechanical energy: ∆K + ∆U = 0. K = -∆U = -mg∆y The (translational) kinetic energy of the block increases from 0 to Kb=12mv2. The (rotational) kinetic energy Kp of the pulley increases from 0 to Kp=12Iω2. The total of these KE's is (Kb + Kp) - (0+0) = -mg∆y, so Kb + Kp = -mg∆y. Rim speed = block speed v = rω In terms of R and pulley mass mp, I = 1/2mpR², so Kp = 1/2 (1/2mpR²)(v/R)². Final equation: Kp = 1/4(mpv²). Use v = rw to solve for w, with new v value for v.

1D explosion: Hauler moving along an x axis in space, has INTERNAL explosion, leaves a module behind. Given initial V relative to sun, masses, V hauler relative to module. What is V hauler relative to sun?

Pi = Mvi

Angular velocity derived from angular position: Disk rotating around its axis like a merry go round. What is the angular position of the disk at t= −2.00 s? Theta(t) equation is given.

Plug time into equation.

Convert radians to degrees

Rad x 180/pi = degrees

The figure shows the position vector of a particle at a certain instant, and four choices for the direction of a force that is to accelerate the particle, with all four choices in the xy plane. Rank the choices according to the magnitude of the time rate of change dℓ/dt they produce in the angular momentum of the particle about point O.

Rank the forces according to the magnitudes of the torque they produce about point O, by using τ=rF⊥. A greater torque magnitude τ means a greater magnitude of dℓ/dt. Forces F→2 and F→4 do not have a perpendicular component F⊥ and thus cannot change the angular momentum.

For three situations, here are values for the grindstone's initial angular velocity ω0 and constant angular acceleration α. Will the grindstone come to a momentary stop?

Same sign = will not stop, different signs = will stop.

A group of six circus penguins has asked you to help them with a balancing act. Half of them are to be on one side of a traditional seesaw and half on the other side. A sketch of the seesaw is shown below, including the positions at which the penguins are allowed to sit, marked off in meters from the pivot at the center. Penguin B knows to sit at the 4 m position on the left, and penguin F knows to sit at the 4 m position on the right. Where should the rest sit?. Here are their masses.

Set both sides equal with those on the right with a negative sign. Plug in values of each location to see where both sides = zero.

Here are two arrangements of particles that have the same mass and speed. In this first figure, particles 1 and 2 move around point O in circles with radii 2 m and 4 m. In the second figure, particles 3 and 4 travel along straight lines at perpendicular distances of 4 m and 2 m from point O. Particle 5 moves directly away from O. Rank the five particles according to the magnitudes of their angular momentum about point O. Indicate the ranking by moving the labels into the boxes below,

Since mass and velocity are the same, use l = mrv and plug in different radii. Magnitude = use absolute value. Use right hand rule to determine (+) or (-) by pushing radius into force vector.

A tomahawk expert knows how to throw the instrument so that it completes an integer number of full revolutions about its center of mass during its flight, to bury its edge in the target. Suppose that for a flight of horizontal distance d = 5.90 m, with a horizontal component of velocity vx = 20.0 m/s, a tomahawk rotates 1.00 rev. The tomahawk's rotational inertia is Icom = 1.95 × 10−3 kg ⋅ m2. We want the magnitude of its angular momentum about the center of mass during the flight. What is the angular momentum L of the tomahawk in flight?

Since you solved for w, I is given with Icom statement.

Newton's second law and constant angular-acceleration equations: The figure shows a 4.0 kg disk that can turn about its center like a merry-go-round. Three forces act on the rim at radius R = 0.20 m, starting at time t = 0. The disk begins at rest but the torques associated with the forces cause it to rotate. During the rotation, the forces maintain their orientations with respect to the disk. We want the angular displacement θ − θ0 of the disk at t = 5.0 s. What is the angular acceleration α of the disk?

Solve for all torques and net torque using equation t = Nsinangle. Recall that for a uniform disk rotating around its central axis, I=12MR2. Plug I value into equation Tnet = I alpha.

If the record were to rotate in the opposite sense, what would be the direction of the angular velocity vector?

Up the z axis, while clockwise would be down the Z axis. Curl fingers in the direction the album is rotating, remembering to ONLY USE your R hand.

Rotational energy, spin test explosion: There was spin testing of a sample of a solid steel rotor (disk) of mass M = 272 kg and radius R = 38.0 cm. When the sample reached angular speed w of 14,000 rev/min, explosion occurred. How much energy was released in the explosion?

Use I = MR^2 to find I, find w with w equation, plug into K = 1/2w^2. Make sure to convert min to s

Motion of the com along three particles on a graph

Use acom equation for both x and y, make sure to add VECTORS, don't just add algebraically. The two values will give you two legs of the triangle. Need to use pythagorean theorem to find the actual acceleration vector. Use the comx and comy values to find missing angle (use inverse tangent)

Newton's second law and constant angular-acceleration equations: The figure shows a 4.0 kg disk that can turn about its center like a merry-go-round. Three forces act on the rim at radius R = 0.20 m, starting at time t = 0. The disk begins at rest but the torques associated with the forces cause it to rotate. During the rotation, the forces maintain their orientations with respect to the disk. We want the angular displacement θ − θ0 of the disk at t = 5.0 s. What is the angular displacement θ-θ0 of the disk from time t= 0 to t= 5.0 s?

Use constant acceleration equations, taking advantage of the fact that w0 = 0 since the object started at rest.

Figure shows a uniform disk with mass M = 2.5 kg and radius R = 20 cm, mounted on a fixed horizontal axle. A block with mass m= 1.2 kg hangs form a massless cord that is wrapped around the rim of the disk Find the tension in the cord.

Use equation T = -1/2 Ma

A grindstone rotates at constant angular acceleration α=0.35 rad/s2. At time t=0, it has an angular velocity of ω0= -4.6 rad/s and a reference line on it is horizontal, at the angular position θ0 = 0. a. At what time after t=0 is the reference line at the position θ=5.0rev?

Use equation θ - θ₀ = ω₀t + 1/2αt². Make sure to convert revs by multiplying by 2π (1 revolution).

Find the center of mass of three plates on an xy coordinate graph.

Xcom = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3) Do the same for Ycom Use equation d = m/v, rearrange for m, to get m = ptA (where p is the term "rho" for density) A will equal x times y of each square.

Linear and angular variables, roller coaster speed up: As passenger accelerates along the arc, the magnitude of this centripetal acceleration increases alarmingly. When the magnitude "a" of the net acceleration reaches 4g at some point P and angle thetaP along the arc, we want the passenger to move in a straight line, along a tangent to the arc. What is the magnitude a of the passenger's net acceleration at point P and after point P?

a = 4g before point P, a = g after point p since they now move in a straight line and there is no centripetal acceleration.

A cockroach rides the rim of a merry-go-round. (a) If the angular speed ω is constant, does the cockroach have radial acceleration? (b) If ω is constant, does the cockroach have tangential acceleration? (c) If ω is increasing, does the cockroach have radial acceleration? (d) If ω is increasing, does the cockroach have tangential acceleration?

a) yes- always when moving in a circle b) no c) yes d) yes- only changes when w is increasing

Linear and angular variables, roller coaster speed up: As passenger accelerates along the arc, the magnitude of this centripetal acceleration increases alarmingly. When the magnitude "a" of the net acceleration reaches 4g at some point P and angle thetaP along the arc, we want the passenger to move in a straight line, along a tangent to the arc. What angle thetaP should the ac subtend so that "a" is 4g at point P?

aT and r are constant, so we can use constant acceleration equations.

Conservation of linear momentum equation

mv = m1v1 + m2v2 + m3v3

A tomahawk expert knows how to throw the instrument so that it completes an integer number of full revolutions about its center of mass during its flight, to bury its edge in the target. Suppose that for a flight of horizontal distance d = 5.90 m, with a horizontal component of velocity vx = 20.0 m/s, a tomahawk rotates 1.00 rev. The tomahawk's rotational inertia is Icom = 1.95 × 10−3 kg ⋅ m2. We want the magnitude of its angular momentum about the center of mass during the flight. What is the angular speed ω?

w = theta/time, rearrange simple time equation t=d/v and plug in values.

What does negative w mean? positive w?

Negative = clockwise rotation, positive means counterclockwise.

2D completely inelastic collision: 2 skaters collide and embrace in a completely inelastic collision, origin at point of collision. Given masses, initial velocities. Want final V after collision in magnitude angle notation and velocity com before and after collision.

Solve for Vx and Vy using the given graph. Use momentum conservation equation (see pic). Make a component triangle to find hypotenuse vector V. Use inverse tan to get the angle. SPEED OF COM AND ANGLE remain constant, even before the collision.

2D glancing collision of projectile and initially stationary target. Given projectile's initial momentum, final x component, and final y component. What is target's final x and y component?

Use p1i + p2i = p1f + p2f for both x and y components given by the graph.

A child's top is spun with angular acceleration a = 5t^3-4t with t in seconds and in radians per second-squared. At t=0, the top has angular velocity 5 rad/s, and a reference line on it is at angular position = 2 rad. Obtain expression for angular velocity w(t) of the top.

a(t) is the derivative of w(t) with respect to time. Find w(t) by integrating a(t) with respect to time.

A tall, cylindrical chimney will fall over when its base is ruptured. Treat chimney as a thin rod of length L = 55 m. At the instant it makes an angle of 35 degrees with the vertical, what is its angular speed wf?

-Energy is conserved, sum of K and U do not change. -Rotational kinetic energy = 1/2 Iω² -Kf + Uf = Ki + Ui -Icom is 1/12 mL² + mh² but chimney rotates around 1 end at distance L/2 from center, so use the parallel axis theorem to get I = 1/12 mL² + m(L/2)² = 1/3 mL². -Substitute this into K = 1/2 Iω² to get Kf= 1/2(1/3mL²)ω². -Potential energy: assume all mass concentrated at com, which is at heigh 1/2L. Ui = 1/2mgL -When chimney has rotated through angle θ, height is 1/2Lcosθ. PE is now Uf = 1/2mgLcosθ. -Ki is zero because it wasn't moving originally -Substitute all these into Kf + Uf = Ki + Ui and solve for ω.