Doc Practice Questions (MCQs & FRQs) Torque and Rotational Motion - AP

SECTION I: MULTIPLE CHOICE

Directions

Answer the following 20 multiple-choice questions. Each question has four answer choices (A, B, C, D). Select the best answer for each question. You may use a calculator where appropriate.

Questions 1-2 refer to the following scenario:

1. Which of the following equations correctly represents the torque balance about the hinge for this system?

1. \( T(3.0) = Mg(2.0) + mg(4.0) \)
2. \( T(3.0) = Mg(4.0) + mg(2.0) \)
3. \( T(4.0) = Mg(2.0) + mg(3.0) \)
4. \( T(2.0) = Mg(3.0) + mg(4.0) \)

2. If the cable were moved closer to the hinge while maintaining equilibrium, what would happen to the tension in the cable?

1. The tension would decrease because the lever arm decreases.
2. The tension would increase because the lever arm decreases.
3. The tension would remain the same because the total weight hasn't changed.
4. The tension would decrease because the angle of the cable changes.

3. A solid disk and a hollow ring, both of mass M and radius R, are released simultaneously from rest at the top of an inclined plane. Both objects roll without slipping. Which statement correctly describes the motion?

1. The disk reaches the bottom first because it has a smaller rotational inertia.
2. The ring reaches the bottom first because more of its mass is far from the rotation axis.
3. Both reach the bottom simultaneously because they have the same mass.
4. Both reach the bottom simultaneously because they have the same radius.

4. What is the magnitude of the net torque on the door about the hinge axis?

1. 50 N·m
2. 60 N·m
3. 72 N·m
4. 120 N·m

5. A thin rod of length L and mass M is pivoted about one end. The rotational inertia about this pivot point is \( I = \frac{1}{3}ML^2 \). If the same rod were pivoted about its center instead, what would be the rotational inertia about the new pivot point?

1. \( \frac{1}{12}ML^2 \)
2. \( \frac{1}{6}ML^2 \)
3. \( \frac{1}{4}ML^2 \)
4. \( \frac{1}{3}ML^2 \)

Data Table 1: A student measures the angular position θ (in radians) of a rotating wheel as a function of time t (in seconds):

6. Based on the data in Table 1, which of the following best describes the angular motion of the wheel?

1. The wheel rotates with constant angular velocity of 2.0 rad/s.
2. The wheel rotates with constant angular acceleration of 2.0 rad/s².
3. The wheel rotates with constant angular acceleration of 4.0 rad/s².
4. The wheel rotates with increasing angular acceleration.

7. A figure skater is spinning with arms extended. When the skater pulls their arms inward, which of the following correctly describes what happens and why?

1. Angular velocity increases because angular momentum is conserved and rotational inertia decreases.
2. Angular velocity decreases because angular momentum is conserved and rotational inertia decreases.
3. Angular velocity increases because rotational kinetic energy is conserved.
4. Angular velocity remains constant because no external torque is applied.

8. A uniform disk of radius R rolls without slipping on a horizontal surface with linear speed v of its center of mass. What is the speed of a point on the top edge of the disk?

1. 0
2. v
3. \( \sqrt{2}v \)
4. 2v

9. Which of the following expressions correctly relates the rotational inertia I of the wheel to the measured quantities?

1. \( I = \frac{mgr}{a} \)
2. \( I = \frac{mr(g-a)}{a} \)
3. \( I = \frac{m(g-a)r^2}{a} \)
4. \( I = \frac{mgr^2}{a} \)

10. A thin hoop, a solid disk, and a solid sphere all have the same mass and radius. They are simultaneously released from rest at the top of an inclined plane and roll without slipping. Which object reaches the bottom first?

1. The hoop, because it has the largest rotational inertia
2. The disk, because it has intermediate rotational inertia
3. The sphere, because it has the smallest rotational inertia
4. All three reach the bottom simultaneously

11. What physical quantity does the slope of the line in Graph 1 represent?

1. Angular displacement
2. Angular acceleration
3. Rotational inertia
4. Net torque

12. Using the information from Graph 1, what is the total angular displacement of the disk during the 4-second interval?

1. 20 rad
2. 40 rad
3. 60 rad
4. 80 rad

13. A uniform meter stick is supported at the 40 cm mark. A 200 g mass is hung at the 10 cm mark. Where should a 300 g mass be placed to balance the meter stick? (Assume the meter stick has negligible mass.)

1. 50 cm mark
2. 60 cm mark
3. 70 cm mark
4. 80 cm mark

14. A disk rotates about a fixed axis with constant angular acceleration. If the disk starts from rest and reaches an angular velocity of 12 rad/s after 3.0 seconds, how many complete revolutions does it make during this time?

1. 1.4 revolutions
2. 2.9 revolutions
3. 5.7 revolutions
4. 18 revolutions

15. If the door remains stationary, what must be true about the magnitudes of the forces?

1. Student A applies twice the force that Student B applies.
2. Student B applies twice the force that Student A applies.
3. Both students apply equal forces.
4. The ratio depends on the mass of the door.

16. A solid cylinder of mass M and radius R rolls without slipping down an incline of height h. What is the linear speed of the cylinder's center of mass at the bottom of the incline?

1. \( \sqrt{gh} \)
2. \( \sqrt{\frac{3gh}{2}} \)
3. \( \sqrt{\frac{4gh}{3}} \)
4. \( \sqrt{2gh} \)

17. A rotating platform has rotational inertia I and is spinning with angular velocity ω. A person standing at the center walks outward to the edge. Assuming no external torques act on the system, which statement is correct?

1. The angular velocity increases because the person's linear velocity increases.
2. The angular velocity decreases because the system's rotational inertia increases.
3. The angular velocity remains constant because angular momentum is conserved.
4. The angular velocity decreases because the person exerts a torque on the platform.

Data Table 2: A student collects data for a rotating object with rotational inertia I = 0.40 kg·m² subjected to various net torques τ:

18. The data in Table 2 best support which relationship?

1. τ = Iα, confirming Newton's second law for rotation
2. τ = Iω, showing torque is proportional to angular velocity
3. α = τ/r, where r is the radius
4. τ = L/t, where L is angular momentum

19. A massless rod of length 2L has two point masses attached: mass m at one end and mass 2m at the other end. The rod rotates about an axis perpendicular to the rod through its center. What is the rotational inertia of this system?

1. 2mL²
2. 3mL²
3. 4mL²
4. 6mL²

20. A wheel initially at rest undergoes constant angular acceleration. After 5.0 seconds, it has completed 25 revolutions. What is the magnitude of its angular acceleration?

1. 2.0 rad/s²
2. 4.0 rad/s²
3. 6.3 rad/s²
4. 10 rad/s²

SECTION II: FREE RESPONSE

Directions

Answer both of the following free-response questions. Show all your work clearly. For calculations, you must show all steps including formulas used, algebraic manipulation, and substitution of values with units. For experimental design and analysis questions, provide complete, detailed responses using appropriate physics terminology.

Suggested time: 50 minutes (25 minutes per question)

FRQ 1: Mathematical Routines (25 minutes)

1. Begin your derivation by writing a fundamental physics principle or equation. Derive an expression for the linear acceleration a of the sphere's center of mass as it rolls down the incline. Express your answer in terms of g, θ, and numerical constants. Show all steps in your derivation.

2. Derive an expression for the minimum coefficient of static friction μ_s between the sphere and the incline that is necessary to prevent slipping. Express your answer in terms of θ and numerical constants.

3. A solid cylinder with the same mass M and radius R is released from the same height on the same incline. The rotational inertia of a solid cylinder is \( I = \frac{1}{2}MR^2 \). Determine the ratio of the time it takes for the sphere to reach the bottom to the time it takes for the cylinder to reach the bottom. Show your reasoning.

4. The sphere rolls down an incline of length d = 2.0 m and angle θ = 30°. Calculate the linear speed of the sphere's center of mass when it reaches the bottom of the incline. Express your answer numerically with appropriate units.

FRQ 2: Experimental Design and Analysis (25 minutes)

1. Describe a detailed experimental procedure the student should follow to collect data that can be used to determine the rotational inertia of the wheel. In your description, include:

   • What quantities should be measured
   • How each quantity should be measured using the available equipment
   • What measurements should be varied to collect multiple data points

2. Explain what the student should graph to produce a linear relationship. Identify what quantities should be plotted on each axis, and state what physical quantity the slope of the best-fit line represents. Justify your answer with appropriate physics reasoning or equations.

3. The student performs the experiment and collects the following data:

   

   The axle radius is r = 0.040 m. Complete the third column of the data table by calculating 1/a for each trial.

4. On the grid provided (or describe what you would plot), graph the appropriate quantities to produce a linear relationship. Draw a best-fit line through your data points.

   [Note: In an actual exam, graph paper would be provided. For this practice, you may sketch your graph on separate paper.]

5. Using your graph, calculate the rotational inertia I of the bicycle wheel. Show all work including how you determined the slope and the subsequent calculation.

ANSWER KEY

Part A: Multiple Choice Answer Table

Part B: Detailed FRQ Answers

FRQ 1 - Answer Key

Part A: Derive linear acceleration

Step 1: Begin with Newton's second law for translation and rotation.

For translation (down the incline):
\( Mg\sin\theta - f = Ma \)

For rotation about the center of mass:
\( \tau = I\alpha \)
\( fr = I\alpha \)

Step 2: Apply the rolling without slipping condition.

\( a = r\alpha \)
Therefore: \( \alpha = \frac{a}{r} \)

Step 3: Substitute into the rotational equation.

\( fr = I\left(\frac{a}{r}\right) \)
\( f = \frac{Ia}{r^2} \)

Step 4: Substitute the expression for friction into the translational equation.

\( Mg\sin\theta - \frac{Ia}{r^2} = Ma \)

Step 5: Substitute the rotational inertia of a solid sphere.

\( Mg\sin\theta - \frac{\left(\frac{2}{5}MR^2\right)a}{R^2} = Ma \)
\( Mg\sin\theta - \frac{2}{5}Ma = Ma \)

Step 6: Solve for a.

\( Mg\sin\theta = Ma + \frac{2}{5}Ma \)
\( Mg\sin\theta = Ma\left(1 + \frac{2}{5}\right) \)
\( Mg\sin\theta = Ma\left(\frac{7}{5}\right) \)
\( a = \frac{5g\sin\theta}{7} \)

Final Answer: \( a = \frac{5g\sin\theta}{7} \)

Part B: Derive minimum coefficient of static friction

Step 1: The friction force required for rolling without slipping is found from Part A.

\( f = \frac{Ia}{r^2} = \frac{\left(\frac{2}{5}MR^2\right)\left(\frac{5g\sin\theta}{7}\right)}{R^2} \)
\( f = \frac{2Mg\sin\theta}{7} \)

Step 2: The maximum static friction available is:

\( f_{max} = \mu_s N \)
where \( N = Mg\cos\theta \) (normal force on incline)

Step 3: For rolling without slipping, we need \( f \leq f_{max} \):

\( \frac{2Mg\sin\theta}{7} \leq \mu_s Mg\cos\theta \)

Step 4: Solve for μ_s.

\( \mu_s \geq \frac{2\sin\theta}{7\cos\theta} \)
\( \mu_s \geq \frac{2\tan\theta}{7} \)

Final Answer: The minimum coefficient of static friction is \( \mu_s = \frac{2\tan\theta}{7} \)

Part C: Determine time ratio

Step 1: For the solid sphere, we found \( a_{sphere} = \frac{5g\sin\theta}{7} \)

Step 2: For the solid cylinder, repeat the derivation with \( I = \frac{1}{2}MR^2 \):

\( Mg\sin\theta - \frac{\left(\frac{1}{2}MR^2\right)a}{R^2} = Ma \)
\( Mg\sin\theta - \frac{1}{2}Ma = Ma \)
\( Mg\sin\theta = \frac{3}{2}Ma \)
\( a_{cylinder} = \frac{2g\sin\theta}{3} \)

Step 3: Use kinematic equation \( d = \frac{1}{2}at^2 \) (starting from rest).

For the same distance d:
\( t = \sqrt{\frac{2d}{a}} \)

Step 4: Find the ratio of times.

\( \frac{t_{sphere}}{t_{cylinder}} = \sqrt{\frac{a_{cylinder}}{a_{sphere}}} = \sqrt{\frac{\frac{2g\sin\theta}{3}}{\frac{5g\sin\theta}{7}}} = \sqrt{\frac{2 \times 7}{3 \times 5}} = \sqrt{\frac{14}{15}} \)

Final Answer: \( \frac{t_{sphere}}{t_{cylinder}} = \sqrt{\frac{14}{15}} \approx 0.966 \)
The sphere reaches the bottom in slightly less time than the cylinder because it has smaller rotational inertia, allowing more of its gravitational potential energy to convert to translational kinetic energy.

Part D: Calculate speed numerically

Step 1: Use the kinematic equation \( v^2 = v_0^2 + 2ad \) with \( v_0 = 0 \).

\( v^2 = 2ad \)

Step 2: Substitute known values.

\( a = \frac{5g\sin\theta}{7} = \frac{5(9.8\text{ m/s}^2)\sin(30°)}{7} = \frac{5(9.8)(0.5)}{7} = \frac{24.5}{7} = 3.5\text{ m/s}^2 \)

\( v^2 = 2(3.5\text{ m/s}^2)(2.0\text{ m}) = 14\text{ m}^2\text{/s}^2 \)

Step 3: Solve for v.

\( v = \sqrt{14} = 3.7\text{ m/s} \)

Final Answer: v = 3.7 m/s

FRQ 2 - Answer Key

Part A: Experimental procedure

A complete procedure should include the following elements:

Measurements to be taken:

• The radius r of the axle (using the meter stick)
• The mass m of each hanging mass (given or measured with a balance)
• The linear acceleration a of the falling mass (using the motion sensor or by measuring distance and time)

Procedure:

1. Measure and record the radius r of the axle using the meter stick.
2. Wrap the string around the axle several times and attach a hanging mass m to the free end.
3. Position the motion sensor below the hanging mass to track its vertical position.
4. Release the mass from rest and use the motion sensor and computer software to record position versus time data.
5. Use the software to determine the linear acceleration a of the falling mass (from the curvature of the position-time graph or from a velocity-time graph).
6. Repeat steps 2-5 with at least four different hanging masses to collect multiple data points.
7. Record all data in an organized table.

What to vary: The mass m of the hanging object should be varied across multiple trials while keeping the wheel and axle radius constant.

Part B: Graph and slope interpretation

To produce a linear relationship, the student should derive the theoretical relationship first.

Derivation:

For the falling mass:
\( mg - T = ma \) ... (1)

For the rotating wheel:
\( \tau = I\alpha \)
\( Tr = I\alpha \)
Since \( a = r\alpha \): \( T = \frac{Ia}{r^2} \) ... (2)

Substitute (2) into (1):
\( mg - \frac{Ia}{r^2} = ma \)
\( mg = ma + \frac{Ia}{r^2} \)
\( mg = a\left(m + \frac{I}{r^2}\right) \)

Rearranging:
\( \frac{1}{a} = \frac{m}{mg} + \frac{I}{mgr^2} \)
\( \frac{1}{a} = \frac{1}{g} + \frac{I}{gr^2} \cdot \frac{1}{m} \)

Graph: Plot 1/a on the vertical axis versus 1/m on the horizontal axis.

Slope interpretation: The slope of the best-fit line equals \( \frac{I}{gr^2} \), which can be used to solve for the rotational inertia I.

The y-intercept equals \( \frac{1}{g} \).

Part C: Complete data table

Part D: Graph construction

Students should create a graph with:

• Horizontal axis: 1/m (in kg^-1), ranging from 0 to approximately 10 kg^-1
• Vertical axis: 1/a (in s²/m), ranging from 0 to approximately 1.0 s²/m

Data points to plot (after calculating 1/m):

• (10.0 kg^-1, 0.80 s²/m)
• (6.67 kg^-1, 0.60 s²/m)
• (5.00 kg^-1, 0.50 s²/m)
• (4.00 kg^-1, 0.44 s²/m)
• (3.33 kg^-1, 0.40 s²/m)

A best-fit line should be drawn through these points. The line should be straight and pass close to all data points.

Part E: Calculate rotational inertia

Step 1: Determine the slope from the graph.

Using two points on the best-fit line (for example, the first and last data points):
Slope = \( \frac{\Delta(1/a)}{\Delta(1/m)} = \frac{0.80 - 0.40}{10.0 - 3.33} = \frac{0.40}{6.67} = 0.060\text{ s}^2\text{·kg/m} \)

Step 2: Use the relationship between slope and rotational inertia.

Slope = \( \frac{I}{gr^2} \)

Therefore:
\( I = \text{slope} \times g \times r^2 \)

Step 3: Substitute known values.

\( I = (0.060\text{ s}^2\text{·kg/m})(9.8\text{ m/s}^2)(0.040\text{ m})^2 \)
\( I = (0.060)(9.8)(0.0016)\text{ kg·m}^2 \)
\( I = 0.00094\text{ kg·m}^2 \)
\( I = 9.4 \times 10^{-4}\text{ kg·m}^2 \)

Final Answer: I = 9.4 × 10^-4 kg·m² or 0.00094 kg·m²

This value represents the rotational inertia of the bicycle wheel about its rotation axis.