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AP Physics C: Mechanics : Circular and Rotational Motion

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Example Question #61 : Ap Physics C

A $2\:kg$ meter stick is nailed to a table at one end and is free to rotate in a horizontal plane parallel to the top of the table. A force of $16\: N$ is applied perpendicularly to its length at a distance $0.67\: m$ from the nailed end.

Calculate the resulting angular acceleration experienced by the meter stick.

Possible Answers:

$16\: rad/s^{2}$

$24\: rad/s^{2}$

$64\: rad/s^{2}$

$7.9\: rad/s^{2}$

Correct answer:

$16\: rad/s^{2}$

Explanation:

For a net torque applied to an object free to rotate,

$\tau _{net}=I\alpha$

The net torque for this problem is supplied by the $16\: N$ force applied at a distance $0.67\: m$ from the pivot. The torque of a force is calculated by

$\tau =Fr\sin\, \theta$

The moment of inertia for a long uniform thin rod about its end is

$I_{rod,end}=\frac{1}{3}mL^{2}$

Combining above,

$Fr\sin90^{\circ}=(\frac{1}{3}mL^{2})\alpha$

Solving for the angular acceleration,

$\alpha =\frac{Fr\sin\, 90^{\circ}}{\frac{1}{3}mL^{2}}=\frac{16N\cdot0.67\: m\cdot1}{\frac{1}{3}\cdot2\, kg\cdot(1\:m)^{2}}=16\: rad/s^{2}$

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Example Question #61 : Motion

A $2\:kg$ meter stick is secured at one end so that it is free to rotate in a vertical circle. It is held perfectly horizontally, and released.

Calculate the instantaneous angular acceleration the moment it is released.

Possible Answers:

$30\: rad/s^{2}$

$60\: rad/s^{2}$

$45\: rad/s^{2}$

$15\: rad/s^{2}$

Correct answer:

$15\: rad/s^{2}$

Explanation:

For a net torque applied to an object free to rotate,

$\tau _{net}=I\alpha$

The net torque for this problem is supplied by the $16\: N$ force applied at a distance $0.67\: m$ from the pivot. The torque of a force is calculated by

$\tau =Fr\sin\, \theta$

The moment of inertia for a long uniform thin rod about its end is

$I_{rod,end}=\frac{1}{3}mL^{2}$

Combining above,

$Fr\sin90^{\circ}=(\frac{1}{3}mL^{2})\alpha$

where the force causing the torque is the meter stick's weight, which is applied at the center of mass of the meter stick - a distance of half the length away from its end.

Solving for the angular acceleration,

$\alpha =\frac{Fr\sin\, 90^{\circ}}{\frac{1}{3}mL^{2}}=\frac{W\cdot\frac{1}{2}L\cdot \sin\, 90^{\circ}}{\frac{1}{3}mL^{2}}$

Solving numerically,

$\alpha =\frac{(2\: kg\cdot 9.8\: m/s^{2})\cdot 0.5\: m\cdot 1}{\frac{1}{3}\cdot 2\, kg\cdot (1\:m)^{2}}=15\: rad/s^{2}$

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Example Question #61 : Mechanics Exam

A 0.18 m long wrench is used to turn the nut on the end of a bolt. A force of 85 N is applied downward to the end of the wrench, as shown in the figure. The angle between the force and the handle of the wrench is 65 degrees.

Ps1 wrench

What is the magnitude and direction of the torque (around the center of the bolt) due to this force?

Possible Answers:

$6.5\: N\cdot m\;downward$

$14\: N\cdot m\;into\;the\;page$

$14\: N\cdot m\;out\;of\;the\;page$

$6.5\: N\cdot m\;out\;of\;the\;page$

$14\: N\cdot m\;downward$

Correct answer:

$14\: N\cdot m\;out\;of\;the\;page$

Explanation:

To calculate the magnitude of the torque,

$\tau =rF\sin\theta$

where the radius is the distance between the center of rotation and the location of the force and the angle $\theta$ is between the radius and the force .

The magnitude is thus,

$\tau =(0.18\: m) \cdot(85\: N)\cdot\sin(65^{\circ})=14\: N\cdot m$

The direction of torque is perpendicular to the plane of the radius and the force , and is given by Right Hand Rule by crossing the radius vector into the force vector. For this situation, the radius vector is left and downward and the force is downward, resulting in the direction of the torque out of the page.

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Example Question #61 : Mechanics Exam

A mouse sits at the edge of a $\small 40cm$ diameter record, rotating on a turntable at fifteen revolutions per minute. If the mouse walks straight inwards to a point $\small 10cm$ from the center, what will be its new angular velocity? Assume the mass of the record is negligible.

Possible Answers:

20rpm

30rpm

45rpm

75rpm

60rpm

Correct answer:

60rpm

Explanation:

Relevant equations:

$L = mr^2\omega$

L_i = L_f

Write an expression for the initial angular momentum.

$L_i = mr_i^2\omega_i=m(20cm)^2(15rpm)$

Write an expression for the final angular momentum.

$L_f = mr_f^2\omega_f=m(10cm)^2\omega_f$

Apply conservation of angular momentum, setting these two expressions equal to one another.

$m(20cm)^2(15rpm)=m(10cm)^2\omega_f$

Solve for the final angular velocity.

$w_f=\frac{(20cm)^2(15rpm)}{(10cm)^2}=60rpm$

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Example Question #71 : Ap Physics C

An ice skater begins to spin, starting with his arms spread out as far as possible, parallel to the ice. He pulls his arms into his body, and then raises them completely vertically towards the ceiling. As the skater pulls his arms inward, his angular velocity will __________, and as he raises his arms vertically his angular velocity will __________.

Possible Answers:

increase . . . stay the same

decrease . . . increase

increase . . . decrease

decrease . . . stay the same

Correct answer:

increase . . . stay the same

Explanation:

Through conservation of angular momentum, as moment of inertia decreases, the angular velocity increases. Moment of inertia is dependent on the distribution of the spinning body's mass away from the center of mass—as the skater brings his arms in, he lowers his moment of inertia, thus increasing his angular velocity. If he raises his arms completely vertically from this point, it does not change the radius of mass distributions (from the center of his body), thus maintaining his angular velocity.

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Example Question #71 : Ap Physics C

Sarah spins a ball of mass attached to a string of length around her head with a velocity v_i . If the ball splits in half, losing exactly one-half of its mass instantaneously, what is its new velocity, v_f ?

Possible Answers:

2v_i

$\frac{v_i}{4}$

4v_i

v_i (no change in velocity)

$\frac{v_i}{2}$

Correct answer:

2v_i

Explanation:

By the conservation of angular momentum, the angular momentum , is equal to the product of the mass, angular velocity, and radius (or length of the rope in this case). The equation relating these terms is:

$L_{initial}=m_{i}v_{i}r$

Here, $m_{i}$ is the initial mass, $v_{i}$ is the initial angular velocity, and is the length of the rope, which remains constant. Angular momentum must be conserved, thus:

$L_{initial}=L_{final}$

Substitute.

$m_{i}v_{i}r=m_{f}v_{f}r$

We are given that $m_{f}=\frac{1}{2}m_{i}$ and $v_{f}$ is the final velocity. Plug in and solve.

$m_{i}v_{i}r=\frac{1}{2}m_{i}v_{f}r$

$v_{f}=2v_{i}$

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Example Question #71 : Mechanics Exam

Rotating platform

A child is standing at the center of a frictionless, rotating platform. Both the child and the platform are rotating with and initial angular velocity, $\small \omega_0$ . The child begins to walk slowly toward the edge of the platform. Which quantity will decrease as the child walks?

Possible Answers:

Total angular momentum

Rotational inertia of the disk

Total momentum

Total rotational kinetic energy

Total angular inertia

Correct answer:

Total rotational kinetic energy

Explanation:

Since the platform is frictionless, as the child walks, angular momentum is conserved. However, since there is an $\small \omega ^{2}$ term in the kinetic energy expression, as $\small \omega$ decreased due to the increase of inertia, it affects the energy more, and it decreases. From the reference frame of the disk, the child feels an outward directed force, and thus is doing negative work. This is the same principle that allows ice skaters to increase their spinning speed.

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Example Question #21 : Rotational Motion And Torque

Two solid cylinderical disks have equal radii. The first disk is spinning clockwise at $18\frac{rad}{s}$ and the second disk is spinning counterclockwise at $4\frac{rad}s{}$ . The second disk has a mass three times larger than the first. If both spinning disks are combined to form one disk, they end up rotating at the same angular velocity and same direction. Find this angular velocity after combination.

Possible Answers:

$1.5\frac{rad}{s}$

$4.5\frac{rad}{s}$

$10.5\frac{rad}{s}$

$8.3\frac{rad}{s}$

$2.0\frac{rad}{s}$

Correct answer:

$1.5\frac{rad}{s}$

Explanation:

For the first disk, we have the information below.

$\omega_1=18\ \frac{rad}{s}$

$m_1=\text{mass of 1st disk}$

$R_1=\text{radius of 1st disk}$

$I_1=\frac{1}{2}m_1R_1^2$

For the second disk, we have the information below.

$\omega_2=-4\ \frac{rad}{s}$

(The minus sign indicates counterclockwise while the positive indicates clockwise.)

m_2=3m_1

R_1=R_2

$I_2=\frac{1}{2}m_2R_2^2=\frac{1}{2}(3m_1)R_1^2=\frac{3}{2}m_1R_1^2$

To do this problem, we use conservation of angular momentum. Before the disks are put in contact, the initial total angular momentum is given by the equation below.

$L_i=I_1\omega_1+I_2\omega_2=\frac{1}{2}m_1R_1^2\omega_1+\frac{3}{2}m_1R_1^2\omega_2$

It is just the sum of the angular momentums of each disk. When the disks are combined together, then final angular momentum can be found by the following equation.

$L_f=I_1\omega+I_2\omega=\frac{1}{2}m_1R_1^2\omega+\frac{3}{2}m_1R_1^2\omega=2m_1R_1^2\omega$

Set this equal to the initial angular momentum.

$\frac{1}{2}m_1R_1^2\omega_1+\frac{3}{2}m_1R_1^2\omega_2=2m_1R_1^2\omega$

Simplify and solve for $\omega$ .

$\omega=\frac{1}{4}\omega_1+\frac{3}{4}\omega_2$

$\omega=\frac{1}{4}(18\ \frac{rad}{s})+\frac{3}{4}(-4\ \frac{rad}{s})$

$\omega=1.5\ \frac{rad}{s}$

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Example Question #21 : Circular And Rotational Motion

A piece of space debris is travelling in an elliptical orbit around a planet. At its closest point to the planet, the debris is travelling at $5000\frac{m}{s}$ . When the debris is 10000km from the planet, it travels at $5000\frac{km}{hr}$ . How close does the debris get to the planet in its orbit?

Possible Answers:

3156km

2778km

1145km

5000km

7191km

Correct answer:

2778km

Explanation:

The equation for conservation of angular momentum is:

$mv_{1}r_{1}=mv_{2}r_{2}$

This means that the angular momentum of the object at the two points in its orbit must be the same. Since the mass of the debris does not change, this gives us an equality of the product of the velocity and distance of the debris at any two points of its orbit:

$v_{1}r_{1}=v_{2}r_{2}$

Plug in known values.

$r_{1}5000\frac{m}{s}=10000km \left(5000\frac{km}{hr}\right)$

There is disagreement between units; the velocities are given in both $\frac{m}{s}$ and $\frac{km}{hr}$ . Glance down at the answer choices and note that they are all lengths in kilometers.

$\left(5000\frac{m}{s}\right)\left(\frac{3600s}{1hr}\right)=18\cdot10^6\frac{m}{hr}$

$\left(18\cdot 10^6\frac{m}{hr}\right)\left(\frac{1m}{1000m}\right)=18000\frac{km}{hr}$

Solve for $r_{1}$ :

$r_{1}=\frac{(10000km)(5000\frac{km}{hr})}{18000\frac{km}{hr}}=\frac{50000}{18}km\approx2778km$

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Example Question #1 : Interpreting Rotational Motion And Torque Diagrams

A mechanic is using a wrench to loosen and tighten screws on an engine block and wants to increase the amount of torque he puts on the screws to adjust them more easily. Which of the following steps will not help him to do so?

Possible Answers:

Pulling or pushing at an angle that is more perpendicular to the wrench

Using a longer wrench

Using a wrench of a different material

Increasing the magnitude of the force he puts on the wrench

Pulling or pushing at an angle that is less perpendicular to the wrench

Correct answer:

Pulling or pushing at an angle that is less perpendicular to the wrench

Explanation:

Remember the torque equation:

$\tau = ||r|| * ||F|| * \sin \theta$

Exerting force on the wrench at an angle less perpendicular to the wrench will reduce $\sin \theta$ and thus reduce the torque.

The other answer options will all increase the torque being applied, making the twisting motion easier.

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AP Physics C: Mechanics : Circular and Rotational Motion

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #61 : Ap Physics C

Example Question #61 : Motion

Example Question #61 : Mechanics Exam

Example Question #61 : Mechanics Exam

Example Question #71 : Ap Physics C

Example Question #71 : Ap Physics C

Example Question #71 : Mechanics Exam

Example Question #21 : Rotational Motion And Torque

Example Question #21 : Circular And Rotational Motion

Example Question #1 : Interpreting Rotational Motion And Torque Diagrams

All AP Physics C: Mechanics Resources

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