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Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What is the angular momentum given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time?

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A solid cylinder of mass 2.0 kg and radius 20 cm is rotating counterclockwise around a vertical axis through its center at 600 rev/min. A second solid cylinder of the same mass is rotating clockwise around the same vertical axis at 900 rev/min. If the cylinders couple so that they rotate about the same vertical axis, what is the angular velocity of the combination?

In the conservation of angular momentum equation, the rotation rate appears on both sides so we keep the (rev/min) notation as the angular velocity can be multiplied by a constant to get (rev/min):
L i = −0.04 kg · m 2 ( 300.0 rev / min), L f = 0.08 kg · m 2 f f f f = −150.0 rev / min clockwise

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A boy stands at the center of a platform that is rotating without friction at 1.0 rev/s. The boy holds weights as far from his body as possible. At this position the total moment of inertia of the boy, platform, and weights is 5.0 kg · m 2 . The boy draws the weights in close to his body, thereby decreasing the total moment of inertia to 1.5 kg · m 2 . (a) What is the final angular velocity of the platform? (b) By how much does the rotational kinetic energy increase?

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Eight children, each of mass 40 kg, climb on a small merry-go-round. They position themselves evenly on the outer edge and join hands. The merry-go-round has a radius of 4.0 m and a moment of inertia 1000.0 kg · m 2 . After the merry-go-round is given an angular velocity of 6.0 rev/min, the children walk inward and stop when they are 0.75 m from the axis of rotation. What is the new angular velocity of the merry-go-round? Assume there is negligible frictional torque on the structure.

I 0 ω 0 = I f ω f ,
I 0 = 6120.0 kg · m 2 ,
I f = 1180.0 kg · m 2 ,
ω f = 31.1 rev / min

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A thin meter stick of mass 150 g rotates around an axis perpendicular to the stick’s long axis at an angular velocity of 240 rev/min. What is the angular momentum of the stick if the rotation axis (a) passes through the center of the stick? (b) Passes through one end of the stick?

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A satellite in the shape of a sphere of mass 20,000 kg and radius 5.0 m is spinning about an axis through its center of mass. It has a rotation rate of 8.0 rev/s. Two antennas deploy in the plane of rotation extending from the center of mass of the satellite. Each antenna can be approximated as a rod has mass 200.0 kg and length 7.0 m. What is the new rotation rate of the satellite?

L i = 1.00 × 10 7 kg · m 2 / s ,
I f = 2.025 × 10 5 kg · m 2 ,
ω f = 7.86 rev / s

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A top has moment of inertia 3.2 × 10 −4 kg · m 2 and radius 4.0 cm from the center of mass to the pivot point. If it spins at 20.0 rev/s and is precessing, how many revolutions does it precess in 10.0 s?

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Challenge problems

The truck shown below is initially at rest with solid cylindrical roll of paper sitting on its bed. If the truck moves forward with a uniform acceleration a , what distance s does it move before the paper rolls off its back end? ( Hint : If the roll accelerates forward with a , then is accelerates backward relative to the truck with an acceleration a a . Also, R α = a a .)

A drawing of a flatbed truck on a horizontal road. The truck is accelerating forward with acceleration a. The bed of the truck has a cylinder on it, a distance d from the back end of the bed.

Assume the roll accelerates forward with respect to the ground with an acceleration a . Then it accelerates backwards relative to the truck with an acceleration ( a a ) .
The forces on a cylinder on a horizontal surface are shown. The cylinder has radius R and moment of inertia one half m R squared and is centered on an x y coordinate system that has positive x to the right and positive y up. Force m g acts on the center of the cylinder and points down. Force N points up and acts at the contact point where the cylinder touches the surface. Force f sub s points to the right and acts at the contact point where the cylinder touches the surface.
Also, R α = a a I = 1 2 m R 2 F x = f s = m a ,
τ = f s R = I α = I a a R f s = I R 2 ( a a ) = 1 2 m ( a a ) ,
Solving for a : f s = 1 2 m ( a a ) ; a = a 3 ,
x x 0 = v 0 t + 1 2 a t 2 ; d = 1 3 a t 2 ; t = 3 d a ;
therefore, s = 1.5 d

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A bowling ball of radius 8.5 cm is tossed onto a bowling lane with speed 9.0 m/s. The direction of the toss is to the left, as viewed by the observer, so the bowling ball starts to rotate counterclockwise when in contact with the floor. The coefficient of kinetic friction on the lane is 0.3. (a) What is the time required for the ball to come to the point where it is not slipping? What is the distance d to the point where the ball is rolling without slipping?

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A small ball of mass 0.50 kg is attached by a massless string to a vertical rod that is spinning as shown below. When the rod has an angular velocity of 6.0 rad/s, the string makes an angle of 30 ° with respect to the vertical. (a) If the angular velocity is increased to 10.0 rad/s, what is the new angle of the string? (b) Calculate the initial and final angular momenta of the ball. (c) Can the rod spin fast enough so that the ball is horizontal?

A vertical rod is spinning on its axis. A string is attached to the top of the rod at one end and a ball at the other end. The string hangs down at an angle from the rod.

a. The tension in the string provides the centripetal force such that T sin θ = m r ω 2 . The component of the tension that is vertical opposes the gravitational force such that T cos θ = m g . This gives T = 5.7 N . We solve for r = 0.16 m . This gives the length of the string as r = 0.32 m .
At ω = 10.0 rad / s , there is a new angle, tension, and perpendicular radius to the rod. Dividing the two equations involving the tension to eliminate it, we have sin θ cos θ = ( 0.32 m sin θ ) ω 2 g 1 cos θ = 0.32 m ω 2 g ;
cos θ = 0.31 θ = 72.2 ° ; b. l initial = 0.08 kg · m 2 / s ,
l final = 0.46 kg · m 2 / s ; c. No, the cosine of the angle is inversely proportional to the square of the angular velocity, therefore in order for θ 90 ° , ω . The rod would have to spin infinitely fast.

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A bug flying horizontally at 1.0 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, the stick swings out to a maximum angle of 5.0 ° from the vertical before rotating back. If the mass of the stick is 10 times that of the bug, calculate the length of the stick.

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Practice Key Terms 1

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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