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AP Physics C: Mechanics : Using Torque Equations

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

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Example Questions

Example Question #11 : Rotational Motion And Torque

An art sculpture comprised of a long hollow painted cylinder is designed to pivot about an axis located along its long edge. The cylinder has mass $234\: kg$ , radius $2.2\: m$ , and length $5.8\: m$ .

Calculate the moment of inertia of the sculpture about this axis.

Possible Answers:

$566\: kg\cdot m^{2}$

$2,265\: kg\cdot m^{2}$

$1,130\: kg\cdot m^{2}$

$1,700\: kg\cdot m^{2}$

Correct answer:

$2,265\: kg\cdot m^{2}$

Explanation:

The moment of inertia for a hollow cylinder about its center is given by

$I_{cyl,hollow}=mR^{2}$

The parallel axis is applied here because the pivot axis is located on the outside edge of the cylinder, a distance from the center of mass:

$I=I_{cm}+md^{2}=mR^{2}+mR^{2}=2\cdot mR^{2} = 2\cdot234\: kg\cdot(2.2\: m)^{2}=2,265\: kg\cdot m^{2}$

Incidentally, for this problem, the length does not contribute to the moment of inertia.

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Example Question #11 : Using Torque Equations

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:

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

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

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

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

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 : Ap Physics C

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:

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

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

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

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

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:

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

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

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

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

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

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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AP Physics C: Mechanics : Using Torque Equations

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

All AP Physics C: Mechanics Resources

Example Questions

Example Question #11 : Rotational Motion And Torque

Example Question #11 : Using Torque Equations

Example Question #61 : Ap Physics C

Example Question #61 : Mechanics Exam

All AP Physics C: Mechanics Resources

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