Physics Exam 4
to find speed in m/s take take the sqrt(9.8)(meters given)
A hoop starts from rest at a height 1.5 m above the base of an inclined plane and rolls down under the influence of gravity.What is the linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface? (Neglect friction.)
For tangential acceleration take ∆(ω/t) = 3g / 2.L² = (3*9.8) / 2 = 14.70 rad/s² Convert your large cm to m then multiply (m)(14.70)= m/s^2 For acceleration due to gravity take L' = a / ∆(ω/t) = g / ∆(ω/t) = 9.8m/s² / 14.7 rad/s².. .. ►L' = 0.67m
A meterstick pivoted about a horizontal axis through the 0-cm end is held in a horizontal position and let go.What is the initial tangential acceleration of the 100-cm positionWhich position has a tangential acceleration equal to the acceleration due to gravity?
both translational and rotational kinetic energies.
A rolling cylinder on a level surface has
he cylinder goes higher percentage of difference is 7.1 percent
A uniform sphere and a uniform cylinder with the same mass and radius roll at the same velocity side by side on a level surface without slipping.f the sphere and the cylinder approach an inclined plane and roll up it without slipping, will they be at the same height on the plane when they come to a stop?
answer in meters First find angular displacement by taking (given rev)(2pi/1)= pi radians To find length of arm take given cm value and convert to meters for arc length=(theta)(r) (pi rad)(meters)
At the end of her routine, an ice skater spins through 4.00 revolutions with her arms always fully outstretched at right angles to her body.. If her arms are 60.5 cm long, through what arc length distance do the tips of her fingers move during her finish?
they both have the same energy.
The cans have essentially the same size, shape, and mass. Which can has more energy at the bottom of the ramp? Ignore friction and air resistance.
arc length in meter First find the radius r=m/2= answer in meters For angular distance take ((m/s)/radius )(minutes given)(60 seconds/1) = angular distance in radians For arc length take (radius)(angular distance) = length in meters
The driver of a car sets the cruise control and ties the steering wheel so that the car travels at a uniform speed of 15 m/s in a circle with a diameter of 100 m .Through what angular distance does the car move in 3.50 min ?What arc length does it travel in this time?
answers will be in rad/s All the angular speeds are constant. Then the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/s T = 1 hr = 3600s Then the angular speed of the minute hand is ω =(2π) / (3600s) When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)s Then the angular speed of the hour hand is ω =(2π) / [(12)(3600)s]
What is the angular speed of the second hand of a clock? What is the angular speed of the minute hand of a clock? What is the angular speed of the hour hand of a clock? Are the speeds constant?
look at pictures
A uniform beam 4.45 m long and weighing 2000 N carries a 3350 N weight 1.50 m from the far end, as shown in the figure below (Figure 1). It is supported horizontally by a hinge at the wall and a metal wire at the far end. How strong does the wire have to be? That is, what is the minimum tension it must be able to support without breaking? What are the horizontal component of the force that the hinge exerts on the beam? What are the vertical component of the force that the hinge exerts on the beam?
see picture
Consider a turntable to be a circular disk of moment of inertia It rotating at a constant angular velocity ωi (note that angular velocities use the Greek letter omega and not double-u) around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry") as shown in (Figure 1). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis. Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is Ir. The initial angular velocity of the second disk is zero. There is friction between the two disks. After this "rotational collision," the disks will eventually rotate with the same angular velocity.
L1=WL/w -wL1 WL-wL1 (L2+L3)w/W (1/h)(WLend-wL2-wL3)
Marcel is helping his two children, Jacques and Gilles, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child, Jacques, simply sinking to the ground. Given that Jacques, whose weight is W, is sitting at distance L to the left of the pivot, at what distance L1 should Marcel place Gilles, whose weight is w, to the right of the pivot to balance the seesaw? Find the torque τ about the pivot due to the weight w of Gilles on the seesaw. Determine ∑τ, the sum of the torques on the seesaw. Consider only the torques exerted by the children. Express your answer in terms of W, w, L, and L1. Where should Marcel position Jacques to balance the seesaw?
answer in seconds =v/r =(m/s)/(meters)= rad/s 2pi/(rad/s)= s
The tangential speed of a particle on a rotating wheel is 2.5 m/s . If the particle is 0.22 m from the axis of rotation, how long will the particle take to make one revolution?
Torque by pulley without friction = I *α = (I given)* (acceleration given) = but firctional torque = m*N given so julie need to apply (without friction previous answer)+ (with friction given) = m*N torque to the pulley Convert cm to m For Tension take (previous answer)/( radius)
To start her lawn mower, Julie pulls on a cord that is wrapped around a pulley. The pulley has a moment of inertia about its central axis of I = 0.620 kg⋅m2 and a radius of 5.00 cm . There is an equivalent frictional torque impeding her pull of τf = 0.280 m⋅N . o accelerate the pulley at α = 4.35 rad/s2 , how much torque does Julie need to apply to the pulley? How much tension must the rope exert?
answer in m from left end of board take (kg of painter)(1.5 meters - x) - (kg of board)(2.5meters/2) + N(2.5 meters) = 0. for answer just do 1.1 meters
While standing on a long board resting on a scaffold, a 75-kg painter paints the side of a house, as shown in the figure If the mass of the board is 25 kg , how close to the end can the painter stand without tipping the board over?
lOOK AT picture
he five objects of various masses, each denoted m, all have the same radius. They are all rolling at the same speed as they approach a curved incline. Rank the objects based on the maximum height they reach along the curved incline.
see picture
n the video, the torque due to the mass of the plank is used in the calculations. For this question, ignore the mass of the board. Rank, from largest to smallest, the mass m needed to keep the board from tipping over.
m2 is heavier , so m2 will go down . m1 will go up ; For answer take (m2-m1)(9.8)(meters)-(m*N)/ (mass of pulley)(meters)/2)+(m2+m1)
two masses are suspended from a pulley as shown in the figure (Figure 1). The pulley itself has a mass of 0.10 kg , a radius of 0.20 m , and a constant torque of 0.30 m⋅N due to the friction between the rotating pulley and its axle.What is the magnitude of the acceleration of the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg ? (Neglect the mass of the string.)
2pir
What is the circumference of a circle with a diameter of 10.0 cm?
picture
What is the length of an arc on a circle of radius 30.0 cm when θ=60∘?
The angular acceleration of the pulley is nonzero The cord's tension on the right side of the pulley is higher than on the left side when b moves downward
he figure below shows two blocks suspended by a single cord over a pulley. (Figure 1) The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord's weight can be ignored. Which of the following statements correctly describes the system shown in the figure?
90 cm (10 cm from the weight)
Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot to keep the meter stick in balance?
answer in rad/s^2 for magnitude answer in rev for revolutions ωi=initial angular speed = 0 rad/s ωf rad/sec, t= s , α(angular accelertaion) =(rad/s-0)/seconds= rad/S^2 b)θ( angular displacement)= 0.5(rad/s^2)(seconds)^2= rad No. of revolutions= rad/(2π)= rev
A CD originally at rest reaches an angular speed of 38 rad/s in 5.2 s .What is the magnitude of its angular acceleration? How many revolutions does the CD make in the 5.2 s ?
yes/no Because the distance from the exerted force to the pivot is zero. To find mass first take (centimeters of first given)-( center centimeters) then take (center centimeters)-(0) Finally take (given mass in g)(cm from -0)/ (centimeters from first subtraction) = grams
A centimeter ruler, balanced at its center point, has two coins placed on it, as shown in the figure. (Figure 1)One coin, of mass M1=10g, is placed at the zero mark; the other, of unknown mass M2, is placed at the 4.7 cm mark. The center of the ruler is at the 3.0 cm mark. The ruler is in equilibrium; it is perfectly balanced.Part complete Does the pivot point (i.e., the triangle in the diagram upon which the ruler balances) exert a force on the ruler? Does it exert a nonzero torque about the pivot? Although the pivot exerts a force on the ruler it does not exert a torque with respect to the pivot point. Why not? Find the mass M2.
To find net torque in m*N i=1/2mr^2 i= (.5)(kg)(meters)^2= answer in kg-m^2 Next (rad/s)=(0)+a(seconds)= a in rad/s^2 solve for a Then multiply both answers together
A 2900-kg Ferris wheel accelerates from rest to an angular speed of 0.23 rad/s in 15 s . Approximate the Ferris wheel as a circular disk with a radius of 23 m .What is the net torque on the wheel?
Tension of both in Newtons First add both Newtons together to get gravitational force downward For left take (tool N)(.25)+(shelf N)(.15)/.5 For right take (gravitiaonal force downard)-(resultant from previous)
A 60.0-cm, uniform, 56.0-N shelf is supported horizontally by two vertical wires attached to the sloping ceiling (the figure (Figure 1)). A very small 25.0-N tool is placed on the shelf midway between the points where the wires are attached to it.
answer in rev/s^2 for magnitude answer for time in seconds Initial angular speed w = rev / s = (rev/s)* 2*pi rad / s = rad / s Final angular speed w ' = 0 Angular displacement Q = rev from the relation w'^ 2- w^ 2= 2 aQ the magnitude of the angular acceleration a = [ w'^ 2- w^ 2] / 2Q =[0^ 2- (rev/s)^ 2] / (2*(rev)) = - rev / s^ 2 required time t = [ w' - w]/ a ((rev/s)-(0))/(rev/s)^2
A flywheel rotates with an angular speed of 26 rev/s . As it is brought to rest with a constant acceleration, it turns 52 rev .What is the magnitude of the angular acceleration? How much time does it take to stop?
Bobby has the greater magnitude of linear velocity Both Ana and Bobby have the same magnitude of angular velocity. Both Ana and Bobby have the same magnitude of tangential acceleration. Bobby has the greater magnitude of centripetal acceleration. Both Ana and Bobby have the same magnitude of angular acceleration.
A merry-go-round is rotating at constant angular speed. Two children are riding the merry-go-round: Ana is riding at point A and Bobby is riding at point B. (Figure 1) Which child moves with greater magnitude of linear velocity? Who moves with greater magnitude of angular velocity?Who moves with greater magnitude of tangential acceleration? Who has the greater magnitude of centripetal acceleration?Who moves with greater magnitude of angular acceleration?
First take 2Pi (meters)/seconds given = .5026 m/s angular speed is wf= vf/r = m/s/meters= .rad/s 4(N) ( m) = ( 1/2) ( kg) ( m)^2 a solve for a= rad/s^2 from the rortational kinematic eqaution wf = wi + alpha (t) rad/s = 0 + ( rad/s^2) t solve for t seconds
A planetary space probe is in the shape of a cylinder. To protect it from heat on one side (from the Sun's rays), operators on the Earth put it into a "barbecue mode," that is, they set it rotating about its long axis. To do this, they fire four small rockets mounted tangentially as shown in the figure (Figure 1)(the probe is shown coming toward you). The object is to get the probe to rotate completely once every 50.0 s , starting from no rotation at all. They wish to do this by firing all four rockets for a certain length of time. Each rocket can exert a thrust of 40.0 N . Assume the probe is a uniform solid cylinder with a radius of 4.00 m and a mass of 2000 kg and neglect the mass of each rocket engine.Determine the amount of time the rockets need to be fired.
θ = (laps)x 2π = t = (min)*60=234 sec. ω = Θ/t = laps answer / seconds = answer in rad/s
A race car makes two and a half laps around a circular track in 3.9 min. What is the car's average angular speed?
a)Rotational torque exerted by F1 is τ1 = 12 x0.03 = 0.36 N.m Rotational torque exerted by F2 is τ2 = F2 x0.08= 0.08F2 Equating, F2 = 0.36/0.08 = 4.5 N (b) Nettorque, Στ = 27x0.08 - 12x0.03 = 2.16 - 0.36 = 1.8 N.m Angularacceleration, α = τ/I = 1.8/9 = 0.2 rad/s2 Since, θ = ω1.t + 0.5αt2 ;ω1 = 0 θ = 0.5αt2 t2 = 2θ/α = 2.π/4/0.2 = 7.854 t = √7.854 = 2.8s (c)τ1 = F1.d = 12x0.03 = 0.36N.m τ2 = F3. sin θ xd =F3.x 0.5 x0.08 = 0.04F3 Equating, F3 = 0.36/0.04 = 9.0 N
A rod is bent into an L shape and attached at one point to a pivot. (Figure 1)The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored for this problem. There are three forces that are applied to the rod at different points and angles: F⃗ 1, F⃗ 2, and F⃗ 3. Note that the dimensions of the bent rod are in centimeters in the figure, although the answers are requested in SI units (kilograms, meters, seconds). F3=0 and F1=12N, what does the magnitude of F⃗ 2 have to be for there to be rotational equilibrium? the L-shaped rod has a moment of inertia I=9kgm2, F1=12N, F2=27N, and again F3=0, how long a time t would it take for the object to move through 45∘ ( π/4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Now consider the situation in which F1=12N and F2=0, but now a force with nonzero magnitude F3 is acting on the rod. What does F3 have to be to obtain equilibrium?
No because The torque produce depends on two things i) the force applied and ii) the arm of the moment that is to say the perpendicular distance between the force applied and the point about which the moment acts So the answer to the question you asked is yes a small force can produce a larger torque than a large force if the arm of the moment in the case of the small force is sufficiently large.
A small force and a large force produce torques.Can you tell which one will have the larger torque?
see picture
A square metal plate 0.180 m on each side is pivoted about an axis through point O at its center and perpendicular to the plate. (See the figure below (Figure 1).)Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F1 = 18.0 N, F2 = 27.0 N, and F3 = 14.0 N. The plate and all forces are in the plane of the page.
a= 2g/3r For free body diagram look at picture
A string is wrapped around a uniform solid cylinder of radius r, as shown in (Figure 1). The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass m.Find the magnitude α of the angular acceleration of the cylinder as the block descends.
For normal force of wall equating the torques about the foot of the ladder, we get Newtons*.5*sintheta = N*1*costheta so, N = Newtons <-----answer(1) Normal force of floorl is same as newtons given s Frictional force is the same as normal force of wall
A uniform 280 N ladder rests against a perfectly smooth wall, making a 35.0 ∘ angle with the wall.Find the normal forces that the wall on the ladder.Find the normal forces that the floor exerts on the ladder.What is the friction force on the ladder at the floor?
answer in Newtons T1 = torque by the force T2 = torque by gravity at centre of mass T1 + T2 = 0 T1 = F*sin(theta)*d T2 = - mg*cos(theta)*d /2 tan(theta) = meters tall /meters in depth and width= theta = tan^-1( previous resultant) 0 = T1 +T2 0 =(F*sin(theta)*d) + (- mg*cos(theta)*d/2) F = ( kg*9.8*cos(theta)*meters depth) / (sin(theta)*meters tall) F = N
A worker applies a horizontal force to the top edge of a crate to get it to tip forward (see the figure (Figure 1)).If the create has a mass of 50 kg and is 1.6 m tall and 0.80 m in depth and width, what is the minimum force needed to make the crate start tipping? (Assume the center of mass of the crate is at its center and static friction great enough to prevent slipping.)
angular velocity in rev/s revolutions in rev tangential speed in m/s magnitude of resultant in m/s^2 First take the diameter in m and find the radius then (inital angular velocity)+(acceleration)(time in seconds)= rev/s For revolutions take (0)+(inital angular velocity)(seconds)+(.5)(acceleration)(seconds^2)= rev then for speed take (radius)(rev/s from first part)(2pi)= m/s Finally for magnitude take first take (radius)(acceleration)(2pi)= m/s then take (radius)( rev/s from part one)^2(2pi)= m/s then take sqrt( resultant 1)^2+ (resultant 2)^2= m/s^2
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.290 rev/s . The magnitude of the angular acceleration is 0.912 rev/s2 . Both the the angular velocity and angular accleration are directed clockwise. The electric ceiling fan blades form a circle of diameter 0.800 m .Compute the fan's angular velocity magnitude after time 0.192 s has passed. hrough how many revolutions has the blade turned in the time interval 0.192 s from Part A? What is the tangential speed vtan(t) of a point on the tip of the blade at time t = 0.192 s ? What is the magnitude a of the resultant acceleration of a point on the tip of the blade at time t = 0.192 s ?
acceleration is in rev/s^2 number of revolutions is in rev time answer in seconds
An electric fan is turned off, and its angular velocity decreases uniformly from 550 rev/min to 180 rev/min in 3.50 s . Find the angular acceleration in rev/s2. Find the number of revolutions made by the motor in the 3.50 s interval. How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part A?
To find angular speed take (kg*m^2 given)(rps given)= rps
An ice skater has a moment of inertia of 2.6 kg⋅m2 when his arms are outstretched and a moment of inertia of 1.0 kg⋅m2 when his arms are tucked in close to his chest.f he starts to spin at an angular speed of 1.2 rps (revolutions per second) with his arms outstretched, what will his angular speed be when they are tucked in?
omentum is always conserved. Angular Momentum = moment of inertia * angular velocity (given percentage)-(100) New angular momentum = new moment of inertia * new angular velocity new angular velocity = (m/s given)÷ decimal of previous percentage found = rad/s
An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 25 % What is the resulting angular speed? By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)
Get answer in rad/s^2 for magnitude First take rpm(2pi/60)= rad/s Then take (0-rad/s)/time in sec= (rad/s^2) For revolutions (.5)(rpm)(seconds)/60
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration.If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, what is α, the magnitude of the angular acceleration of the CD, as it spins to a stop? ow many revolutions does the CD make as it spins to a stop?
By circumference = 2 x pi x r =>12 = 2 x 3.14 x r =>r = 1.91 m Area enclosed (A) = pi x r^2 =>A = 3.14 x (1.91)^2 =>A = 11.46 m^2
Suppose that you needed to make a pen for some small animals. You have 12.0 m of fencing. You decide to make a circular pen, because if you wish to enclose an area using a given length of fencing, then a circular fence encloses a larger area than a fence of any other shape. What is the area A of the pen?
answer for magnitude in rad/s^2 answer for revolutions in rev ) ω = rpm turn=rpm x 2π / 60 = rad/s initial ω1 = rpm switched speed = rpm 2π / 60 = rad/s 1 α = ω1 - ω /t = (rpm1-rpm)/seconds=(rad/s2) b) v2 - u2 = 2as Solve for theta in the following (rad/s)1 - (rad/s initial) = 2 x (rad/s^2) x θ theta will be in radians = radians / 2π = revolutions
The blades of a fan running at low speed turn at 230 rpm . When the fan is switched to high speed, the rotation rate increases uniformly to 350 rpm in 5.70 s .What is the magnitude of the angular acceleration of the blades? How many revolutions do the blades go through while the fan is accelerating?
look at picture
The boom in the figure below (Figure 1) weighs 2500 N and is attached to a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 36.5 % of its length.
He spins slower The bean bag is not rotating before it contacts the experimenter, so (by conservation of rotational momentum) he must supply the rotational momentum required to make it spin at the same rate as his lap. Therefore, he ends up spinning slower.
The experimenter from the video rotates on his stool, this time holding his empty hands in his lap. You stand on a desk above him and drop a long, heavy bean bag straight down so that it lands across his lap, in his hands. What happens?
For Tension take First take sintheta= (side meters)/(hyptoneuse) (mass in newtons given)(bottom meters)+ (mass in newtons that's hanging)(bottom meters)/ (sin theta)(bottom meters)= Newtons For horizontal component take costheta=(bottom meters)/(hypotneuse) (newtons from previous)(costheta)= Newtons For vertical component take (mass in newtons given)+(mass hanging in newtons)- ( sin theta from before)( tension from part 1)
The horizontal beam in the figure below (Figure 1) weighs 110 N , and its center of gravity is at its center.Find the tension in the cable.Find the horizontal component of the force exerted on the beam at the wall.Find the vertical component of the force exerted on the beam at the wall.
Length of hour hand l = m Angular speed of hour hand ω = 1.45*10^-4 rad/s time t = (minutes given)*60 s = seconds Distance covered S = lωt = meters * 1.45*10^-4 * seconds found in previous= answer in meters Length of minute hand l = m Angular speed of hour hand ω = 1.75*10^-3 rad/s time t = (minutes given)*60 s = s Distance covered S = lωt = (meters given) * 1.75*10^-3 * (seconds from previous) = answer in meters Length of second hand l = m Angular speed of second hand ω = 0.105 rad/s time t = (minutes given)*60 s = s Distance covered S = lωt = (meters given) * 0.105 * (seconds from previous)= answer in m
The hour, minute, and second hands on a clock are 0.25 m , 0.28 m , and 0.34 m long, respectively. What are the distances traveled by the tips of the hands in a 25 min interval?
see picture
The wrench in the figure has six forces of equal magnitude acting on itRank these forces (A through F) on the basis of the magnitude of the torque they apply to the wrench, measured about an axis centered on the bolt.
The seesaw can be balanced because If the net applied torque on the seesaw is zero then it is said to be in equilibrium and this can be done when the torques applied by the weight of the children are equal. To find distance from pivot point take (kg of child)(meters)/kilograms of playmate = meters
Two children are sitting on opposite ends of a uniform seesaw of negligible mass.Can the seesaw be balanced if the masses of the children are different?If a 34 kg child is 1.3 m from the pivot point (or fulcrum), how far from the pivot point will her 32 kg playmate have to sit on the other side for the seesaw to be in equilibrium?
Well lets see... 20.0 cm2 = 20.0 x 20.0 which they both equal to 400.0cm = area of spoon. 400.0cm x 10 = 4000.0cm
You have a fishing line spool with an end that has an area of 20.0 cm2. How much fishing line do you need to wind around the spool 10 times?
the board with the clown on it f the board is parallel with the horizontal without a clown standing on it, TL = mg(L/2) T = mg/2 T = ( kg of board)(9.8 m/s²) / 2 T = N With the clown, TLcos30° = (m + M)(g)(L/2)cos30° T = (m + M)(g)/2 T = (mass of board + mass of person)(9.8 m/s²) / 2 T = N
n a circus act, a uniform board (length 3.00 m, mass 23.0 kg ) is suspended from a bungie-type rope at one end, and the other end rests on a concrete pillar. When a clown (mass 68.0 kg ) steps out halfway onto the board, the board tilts so the rope end is 30∘ from the horizontal and the rope stays vertical.In which situation will the rope tension be larger? Calculate the force exerted by the rope on the board without the clown on it.Calculate the force exerted by the rope on the board with the clown on it.
