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AP Physics C: Mechanics : Circular and Rotational Motion

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Example Questions

Example Question #11 : Circular And Rotational Motion

A circular disk of radius 0.5m and mass 3kg has a force of 25N exerted perpendicular to its edge, causing it to spin. What is the angular acceleration of the disk?

Possible Answers:

$25\frac{rad}{s^2}$

$40\frac{rad}{s^2}$

$33.3\frac{rad}{s^2}$

$20\frac{rad}{s^2}$

Correct answer:

$33.3\frac{rad}{s^2}$

Explanation:

We can find the angular acceleration using the rotaional motion equivalent of Newton's second law. In rotational motion, torque is the product of moment of inertia and angular acceleration:

$\tau = I\alpha$

The moment of inertia for a circular disk is:

$I=\frac{1}{2}mr^2$

The tourque is the product of force and distance (in this case, the radius):

$\tau=F\cdot r$

We can plug these into our first equation:

$(F)(r)=(\frac{1}{2}mr^2)(\alpha)$

Simplify and rearrange to derive an equation for angular acceleration:

$\alpha=\frac{Fr}{\frac{1}{2}mr^2}=\frac{2F}{mr}$

Use our given values to solve:

$\alpha=\frac{2(25N)}{(3kg)(0.5m)}=33.3\frac{rad}{s^2}$

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Example Question #2 : Using Torque Equations

A 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. Four forces of equal magnitude are applied to the meter stick at different locations. The figure below shows the view of the meter stick from above.

You may assume the forces F_1 and F_2 are applied at the center of the meter stick, and the forces F_3 and F_4 are applied at the end opposite the nail.

Ps1 torques

What is the relationship among the magnitudes of the torques on the meter stick caused by the four different forces?

Possible Answers:

$\tau _{1} = \tau _{2} = \tau _{3} = \tau _{4}$

$\tau _{1} < \tau _{2} = \tau _{3} < \tau _{4}$

$\tau _{1} = \tau _{2} < \tau _{3} = \tau _{4}$

Correct answer:

$\tau _{1} < \tau _{2} = \tau _{3} < \tau _{4}$

Explanation:

Torque is given by,

$\tau =rFsin\theta$

Since all of the forces are equal in magnitude, the magnitude of the torque is then influenced by the radius r and the angle theta between the radius and the force.

For F_1 ,

$\tau_{1} =\left (\frac{1}{2}L \right )F\sin\, 45^{\circ}=0.354\cdot L\cdot F$

For F_2 ,

$\tau_{2} =\left (\frac{1}{2}L \right )F\sin\, 90^{\circ}=0.5\cdot L \cdot F$

For F_3 ,

$\tau_{3} =LF\sin\, 30^{\circ}=0.5 \cdot L \cdot F$

For F_4 ,

$\tau_{4} =LF\sin\, 90^{\circ}=L\cdot F$

Combining this information yields the relationship,

$\tau _{1} < \tau _{2} = \tau _{3} < \tau _{4}$

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Example Question #1 : Using Torque Equations

A man sits on the end of a long uniform metal beam of length $3.05\: m$ . The man has a mass of $85\: kg$ and the beam has a mass of $40\: kg$ .

What is the magnitude of the net torque on the plank about the secured end of the beam? Use gravity $g = 9.8\: m/s^2$ .

Possible Answers:

$3,740\: N\cdot m$

$1,350\: N\cdot m$

$3,140\: N\cdot m$

$1,940\: N\cdot m$

Correct answer:

$3,140\: N\cdot m$

Explanation:

The net torque on the beam is given by addition of the torques caused by the weight of the man and the weight of the beam itself, each at its respective distance from the end of the beam:

$\tau _{net}=\tau _{man}+\tau _{beam}$

Let's assign the direction of positive torque in the direction of the torques of the man's and the beam's weights, noting that they will add together since they both point in the same direction.

$\tau _{net}=L\cdot W_{man}\cdot\sin 90^{\circ}+\left ( \frac{1}{2} L\right )\cdot W_{beam}\cdot\sin\, 90^{\circ}$

$\tau _{net}=L\cdot m_{man}\cdot g+\left ( \frac{1}{2} L\right )\cdot m_{beam}\cdot g$

We can further simplify by combining like terms:

$\tau _{net}=L\cdot g \cdot (m_{man}+\frac{1}{2} m_{beam})$

Using the given numerical values,

$\tau _{net}=(3.05\: m)\cdot(9.8\: m/s^{2})\cdot(85\: kg+\frac{1}{2} 40\: kg)=3140\: N\cdot m$

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Example Question #11 : Circular And Rotational Motion

Two children sit on the opposite sides of a seasaw at a playground, doing so in a way that causes the seasaw to balance perfectly horizontal. The $36\:kg$ child on the left is 1.8 m from the pivot.

What is the mass of the second child if she sits $2.4\: m$ from the pivot?

Possible Answers:

$36\: kg$

$12\: kg$

$48\: kg$

$27\: kg$

Correct answer:

$27\: kg$

Explanation:

A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,

$\tau_{net}=I\alpha$

This is a static situation. There are two torques about the pivot caused by the weights of two children. We will note that these weights cause torques in opposite directions about the pivot, such that

$\tau _{net}=0=\tau _{1}-\tau _{2}$

Consequently,

$\tau _{1}=\tau _{2}$

$r_{1}\cdot m_{1} \cdot g \cdot \sin\: 90^{\circ}=r_{2}\cdot m_{2}\cdot g \cdot \sin\: 90^{\circ}$

Or more simply,

$r_{1}\cdot m_{1}=r_{2} \cdot m_{2}$

Solving for m_2 ,

$m_{2}=\frac{m_{1}r_{1}}{r_{2}}=\frac{36\: kg\cdot1.8\: m}{2.4\: m}}=27\: kg$

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Example Question #1 : Using Torque Equations

A gymnast is practicing her balances on a long narrow plank supported at both ends. Her mass is $55\: kg$ . The $6\:m$ long plank has a mass of $23\: kg$ .

Ps1 gymnast

Calculate the force the right support provides upward if she stands $4\: m$ from the right end. Use gravity $g = 9.8\: m/s^2$ .

Possible Answers:

$380\: N$

$760\: N$

$290\: N$

$880\: N$

$180\: N$

Correct answer:

$290\: N$

Explanation:

A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,

$\tau_{net}=I\alpha$

This is a static situation. As such, any pivot point can be chosen about which to do a torque analysis. The quickest way to the unknown force asked for in the question is to do the torque analysis about the left end of the plank. There are three torques about this pivot: two clockwise caused by the weights of the gymnast and the plank itself, and one counterclockwise caused by the force from the right support. Designating clockwise as positive,

$\tau _{net}=0=\tau _{gymnast}+\tau _{plank}-\tau _{right}$

Consequently,

$\tau _{gymnast}+\tau _{plank}=\tau _{right}$

$r _{gymnast}\cdot W_{gymnast}\cdot\sin\: 90^{\circ}+r _{plank}\cdot W_{plank}\cdot \sin\: 90^{\circ}=r _{right}\cdot F_{right}\cdot \sin\: 90^{\circ}$

This simplifies to

$r _{gymnast}\cdot m_{gymnast}\cdot g+r _{plank}\cdot m_{plank}\cdot g=r _{right}\cdot F_{right}$

Solving for $F_{right}$ ,

$F_{right} = \frac{(r _{gymnast}\cdot m_{gymnast}+r _{plank}\cdot m_{plank})\cdot g}{r _{right}}$

Solving with numerical values,

$F_{right} = \frac{(2\: m\cdot 55\: kg+3\: m\cdot 23\: kg)\cdot 9.8\: m/s^{2}}{6\: m}=290\: N$

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Example Question #1 : Using Torque Equations

A square of side lengths and mass is shown with possible axes of rotation.

Ps1 square

Which statement of relationships among the moments of inertia is correct?

Possible Answers:

I_A = I_B < I_C

I_A = I_B = I_C

I_A < I_B < I_C

I_B < I_A < I_C

Correct answer:

I_A = I_B < I_C

Explanation:

The moments of inertia for both axes and are equal because both of these axes are equivalently passing through the center of the mass.

By the parallel axis theorem for moments of inertia ( $I = I_{cm}+ MD^2$ ), the moment of inertia for axis is larger than or

because it is located a distance

$\frac{L}{2}$ away from the center of mass.

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Example Question #1 : Using Torque Equations

A wind catcher is created by attaching four plastic bowls of mass $0.6\: kg$ each to the ends of four lightweight rods, which are then secured to a central rod that is free to rotate in the wind. The four lightweight rods are of lengths $0.8\: m$ , $0.6\: m$ , $0.4\: m$ , and $0.2\: m$ .

Ps1 wind

Calculate the moment of inertia of the four bowls about the central rod. You may assume to bowls to be point masses.

Possible Answers:

$0.24\: kg \:m^2$

$0.72\: kg\: m^2$

$0.36\: kg\: m^2$

$1.4\: kg\: m^2$

Correct answer:

$0.72\: kg\: m^2$

Explanation:

The moment of inertia for a point mass is I = mR^2 .

To calculate the total moment of inertia, we add the moment of inertia for each part of the object, such that

$I_{total}=I_{1}+I_{2}+I_{3}+I_{4}$

$I_{total}=m_{1}R_{1}^{2}+m_{2}R_{2}^{2}+m_{3}R_{3}^{2}+m_{4}R_{4}^{2}$

The masses of the bowls are all equal in this problem, so this simplifies to

$I_{total}=m (R_{1}^{2}+R_{2}^{2}+R_{3}^{2}+R_{4}^{2})$

Plugging in and solving with numerical values,

$I_{total}=(0.6\: kg) [(0.8\: m)^{2}+(0.6\: m)^{2}+(0.4\: m)^{2}+(0.2\: m)^{2}]=0.72\: kg\cdot m^{2}}$

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Example Question #61 : Ap Physics C

A long uniform thin rod of length $1.82\: m$ has a mass of $12\: kg$ .

Calculate the moment of rotational inertia about an axis perpendicular to its length passing through a point $0.36\: m$ from one of its ends.

Possible Answers:

$40\: kg\cdot m^{2}$

$3.3\: kg\cdot m^{2}$

$4.9\: kg\cdot m^{2}$

$6.9\: kg\cdot m^{2}$

$1.6\: kg\cdot m^{2}$

Correct answer:

$6.9\: kg\cdot m^{2}$

Explanation:

For a long thin rod about its center of mass,

$I_{cm}=\frac{1}{12}mL^{2}$

According to the parallel axis theorem,

$I_{//}=I_{cm}+md^{2}$

where is defined to be the distance between the center of mass of the object and the location of the axis parallel to one through the center of mass. For this problem, is the difference between the given distance and half the length of the rod.

Combining the above,

$I=\frac{1}{12}mL^{2}+m(\frac{1}{2}L-l)^{2}$

Inserting numerical values,

$I=\frac{1}{12}(12)(1.82)^{2}+(12)(\frac{1.82}{2}-.036)^{2}=6.9\: kg\cdot m^{2}$

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Example Question #2 : Using Torque Equations

The moment of inertia of a long thin rod about its end is determined to be $I_0\: kg\: m^2$ .

What is the new value if the mass and length of the rod are both reduced by a factor of ?

Possible Answers:

$I_{new}=\frac{1}{12}I_{0}$

$I_{new}=\frac{1}{4}I_{0}$

$I_{new}=\frac{1}{16}I_{0}$

$I_{new}=\frac{1}{192}I_{0}$

$I_{new}=\frac{1}{64}I_{0}$

Correct answer:

$I_{new}=\frac{1}{64}I_{0}$

Explanation:

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

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

Reducing the mass and the length by a factor of four introduces the following factors into the equation,

$I_{new}=\frac{1}{3}(\frac{1}{4}m)(\frac{1}{4}L)^{2}=\frac{1}{3}(\frac{1}{4})m(\frac{1}{64})L^{2}$

Simplifying,

$I_{new}=(\frac{1}{64})\frac{1}{3}mL^{2}=\frac{1}{64}I_{0}$

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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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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 #11 : Circular And Rotational Motion

Example Question #2 : Using Torque Equations

Example Question #1 : Using Torque Equations

Example Question #11 : Circular And Rotational Motion

Example Question #1 : Using Torque Equations

Example Question #1 : Using Torque Equations

Example Question #1 : Using Torque Equations

Example Question #61 : Ap Physics C

Example Question #2 : Using Torque Equations

Example Question #11 : Rotational Motion And Torque

All AP Physics C: Mechanics Resources

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