AP Physics C Mechanics - 6.2 Torque and Work- Exam Style questions- FRQs

Torque and Work AP  Physics C Mechanics FRQ

Unit 6: Energy and Momentum of Rotating Systems

Weightage : 10-15%

AP Physics C Mechanics Exam Style Questions – All Topics

Question

Object A is a long, thin, uniform rod of mass $M$ and length $2L$ that is free to rotate about a pivot of negligible friction at its left end, as shown above.
(a) Using integral calculus, derive an expression to show that the rotational inertia $I_{A}$ of object A about the pivot is given by $\frac{4}{3}ML^{2}$.
Object B of total mass $M$ is formed by attaching two thin, uniform, identical rods of length $L$ at a right angle to each other. Object B is held in place, as shown above. Express your answers in part (b) in terms of $L$.
(b) Determine the following for the given coordinate system shown in the figure.
i. The x-coordinate of the center of mass of object B
ii. The y-coordinate of the center of mass of object B
Object B has a rotational inertia of $I_{B}$ about its pivot.
(c) Is the value of $I_{B}$ greater than, less than, or equal to $I_{A}$?
_____ Greater than _____ Less than _____ Equal to
Justify your answer.
Object B is released from rest and begins to rotate about its pivot.
(d) On the axes below, sketch graphs of the magnitude of the angular acceleration $\alpha$ and the angular speed $\omega$ of object B as functions of time $t$ from the time it is released to the time its center of mass reaches its lowest point.
(e) While object B rotates from the horizontal position down through the angle $\theta$ shown above, is the magnitude of its angular acceleration increasing, decreasing, or not changing?
_____ Increasing _____ Decreasing _____ Not changing
Justify your answer.
Object B rotates through the position shown above.
(f) Derive an expression for the angular speed of object B when it is in the position shown above. Express your answer in terms of $M$, $L$, $I_{B}$, and physical constants, as appropriate.

Most-appropriate topic codes (AP Physics C: Mechanics):

• Topic 2.1 — Systems and Center of Mass (Part b)
• Topic 5.1 — Rotational Kinematics (Part d)
• Topic 5.4 — Rotational Inertia (Parts a, c)
• Topic 5.6 — Newton’s Second Law in Rotational Form (Parts d, e)
• Topic 6.2 — Torque and Work (Part f)
▶️ Answer/Explanation

(a)
Use the integral definition of rotational inertia:
$I = \int_{r=0}^{r=2L} \lambda r^{2} dr$
$I = \lambda \left[ \frac{r^{3}}{3} \right]_{0}^{2L}$
$I = \frac{\lambda}{3} ((2L)^{3} – 0)$
Substitute the linear mass density $\lambda = \frac{M}{2L}$:
$I = \left( \frac{1}{3} \right) \left( \frac{M}{2L} \right) (8L^{3})$
$I = \frac{4}{3}ML^{2}$

(b)(i)
The x-coordinate of the center of mass of object B is found by treating each arm as a point mass at its center:
$X_{CM} = \frac{\sum m_{i}x_{i}}{\sum m_{i}}$
$X_{CM} = \frac{(\frac{M}{2})(\frac{L}{2}) + (\frac{M}{2})(L)}{\frac{M}{2} + \frac{M}{2}}$
$X_{CM} = \frac{\frac{ML}{4} + \frac{ML}{2}}{M}$
$X_{CM} = \frac{3}{4}L$

(b)(ii)
Similarly, for the y-coordinate:
$Y_{CM} = \frac{\sum m_{i}y_{i}}{\sum m_{i}}$
$Y_{CM} = \frac{(\frac{M}{2})(L) + (\frac{M}{2})(\frac{L}{2})}{\frac{M}{2} + \frac{M}{2}}$
$Y_{CM} = \frac{\frac{ML}{2} + \frac{ML}{4}}{M}$
$Y_{CM} = \frac{3}{4}L$

(c)
Correct choice: Less than
Justification: Because object B has more of its mass distributed closer to the pivot than object A, the rotational inertia of object B must be less than that of object A.

(d)


For the $\alpha$ (angular acceleration) graph: The curve is concave down and begins horizontally with a positive y-intercept.
For the $\omega$ (angular speed) graph: The curve is concave down, starts at the origin ($0$), and ends horizontally at its maximum value.

(e)
Correct choice: Decreasing
Justification: Because the horizontal position of the center of mass for the object is moving closer to the pivot, the lever arm for the downward force of gravity is decreasing. Therefore, the net torque and the resulting angular acceleration decrease.

(f)
Apply conservation of energy from the initial release to the lowest point:
$U_{g1} = K_{2}$
Relate the change in rotational kinetic energy to the change in gravitational potential energy:
$mgh = \frac{1}{2}I\omega^{2}$
Substitute the height change $h = \frac{L}{2}$ (since the center of mass falls by this vertical distance):
$Mg\left(\frac{L}{2}\right) = \frac{1}{2}I_{B}\omega^{2}$
Solve for $\omega$:
$\omega = \sqrt{\frac{2Mg(L/2)}{I_{B}}}$
$\omega = \sqrt{\frac{MgL}{I_{B}}}$

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