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AP Physics C: Mechanics : Mechanics Exam

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

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2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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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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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Example Question #61 : Motion

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:

60rpm

75rpm

20rpm

30rpm

45rpm

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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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Mechanics Exam

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

All AP Physics C: Mechanics Resources

Example Questions

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

Example Question #11 : Using Torque Equations

Example Question #61 : Ap Physics C

Example Question #61 : Mechanics Exam

Example Question #61 : Motion

All AP Physics C: Mechanics Resources

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