AP Physics C Mechanics - 5.3 Torque- Exam Style questions- FRQs

Torque AP Physics C Mechanics FRQ

Unit 5: Torque and Rotational Dynamics

Weightage : 10-15%

AP Physics C Mechanics Exam Style Questions – All Topics

Question

A uniform rod of length $L$ and mass $m$ is attached to a pivot on a vertical pole, as shown in Figure $1$. There is negligible friction between the rod and the pivot. A horizontal string connects Point Q on the rod to the pole. The rod makes an angle $\theta$ with the pole. A block of mass $3m$ hangs from the rod at Point P. The center of mass of the rod is located at Point C.

(a) On the following representation of the rod, draw and label the forces (not components) that are exerted on the rod. Each force must be represented by a distinct arrow that starts on and points away from the point at which the force is exerted on the rod.

(b) In Figure $1$, Point P is located $\frac{3}{8}L$ from the pivot and Point Q is located $\frac{6}{8}L$ from the pivot. Derive an equation for the tension $F_T$ in the horizontal string in terms of $L$, $m$, $\theta$, and physical constants, as appropriate.

(c) The original string is replaced with a longer string that connects Point Q to a higher location on the vertical pole, as shown in Figure $2$. The angle $\theta$ remains the same. How does the new tension $F_{T,new}$ compare with the original tension $F_T$ from part (b)? Justify your reasoning.

(d) A nonuniform rod is now attached to the pivot, as shown in Figure $3$. There is negligible friction between the nonuniform rod and the pivot. The rod has a length of $1.2\text{ m}$ and a linear mass density $\lambda(x)=A+Bx$, where $x$ is the distance from the pivot, $A=6.0\text{ kg/m}$, and $B=10.0\text{ kg/m}^2$.

i. Calculate the mass of the rod.
ii. Calculate the rotational inertia of the rod about the pivot.

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

• Topic 5.3 — Torque (Parts a, b, c)
• Topic 5.4 — Rotational Inertia (Part d)
• Topic 5.5 — Rotational Equilibrium and Newton’s First Law in Rotational Form (Parts a, b, c)

▶️ Answer/Explanation

(a)

The free-body diagram should include four distinct forces originating from their respective points of application. We represent the weight of the rod pulling straight down from the center of mass, and the hanging block pulling straight down at P. The string pulls horizontally to the left at Q, and the pivot exerts a reaction force that generally points upward and to the right to support the system.
1. A downward arrow at Point C labeled $mg$ (force of gravity on the rod).
2. A downward arrow at Point P labeled $3mg$ (tension from the hanging block).
3. A horizontal arrow pointing left at Point Q labeled $F_T$ (tension from the string).
4. An arrow at the pivot pointing upward and rightward labeled $F_{\text{pivot}}$ (reaction force from the hinge).

(b)
To find the tension, we can use the condition for rotational equilibrium, meaning the net torque about the pivot is zero. We sum the clockwise torques from the weights of the rod and the hanging block, and set them equal to the counterclockwise torque from the horizontal string’s tension. By substituting the given distances and lever arms based on angle $\theta$, we solve the equation to isolate the tension force.

$ \sum \tau = 0 $

$ \tau_{\text{string}} = \tau_{\text{rod}} + \tau_{\text{block}} $

$ F_T \left(\frac{6}{8}L \cos\theta\right) = mg \left(\frac{1}{2}L \sin\theta\right) + 3mg \left(\frac{3}{8}L \sin\theta\right) $

$ F_T \left(\frac{3}{4} \cos\theta\right) = mg \left(\frac{4}{8} \sin\theta\right) + mg \left(\frac{9}{8} \sin\theta\right) $

$ F_T \left(\frac{3}{4} \cos\theta\right) = \frac{13}{8} mg \sin\theta $

$ F_T = \frac{13}{8} mg \sin\theta \times \frac{4}{3 \cos\theta} $

$ \boxed{F_T = \frac{13}{6} mg \tan\theta} $

(c)
$ \boxed{F_{T,new} < F_T} $
The new tension $F_{T,new}$ will be less than the original tension $F_{T}$. Because the mass and position of the rod and block remain unchanged, the total clockwise torque due to gravity stays the exact same. However, the new string’s line of action is angled higher, increasing its perpendicular distance (lever arm) from the pivot. Since the required counterclockwise torque remains constant and the lever arm has increased, a correspondingly smaller tension force is needed to maintain equilibrium.

(d)(i)
We find the total mass of the rod by integrating the given linear mass density function over its entire length from $x = 0$ to $x = 1.2\text{ m}$. Setting up the integral $M = \int \lambda(x)\,dx$ and plugging in the given constants for $A$ and $B$, we perform a standard power rule integration. Evaluating this result at the $1.2\text{ m}$ boundary gives the total physical mass of the nonuniform rod.

$ M = \int_{0}^{1.2} (A + Bx) \,dx $

$ M = \left[ Ax + \frac{1}{2}Bx^2 \right]_{0}^{1.2} $

$ M = 6.0(1.2) + \frac{1}{2}(10.0)(1.2)^2 $

$ M = 7.2 + 5(1.44) $

$ M = 7.2 + 7.2 $

$ \boxed{M = 14.4\text{ kg}} $

(d)(ii)
To determine the rotational inertia of the nonuniform rod about the pivot, we integrate the second moment of mass, using the definition $I = \int x^2\,dm$, which expands to $I = \int x^2 \lambda(x)\,dx$. Distributing the $x^2$ into the density function and integrating term by term gives us the polynomial expression for inertia. Finally, plugging in the length bounds yields the final rotational inertia value.

$ I = \int_{0}^{1.2} x^2 (A + Bx) \,dx $

$ I = \int_{0}^{1.2} (Ax^2 + Bx^3) \,dx $

$ I = \left[ \frac{1}{3}Ax^3 + \frac{1}{4}Bx^4 \right]_{0}^{1.2} $

$ I = \frac{1}{3}(6.0)(1.2)^3 + \frac{1}{4}(10.0)(1.2)^4 $

$ I = 2.0(1.728) + 2.5(2.0736) $

$ I = 3.456 + 5.184 $

$ \boxed{I = 8.64\text{ kg}\cdot\text{m}^2} $

AP Physics C Mechanics - 5.3 Torque- Exam Style questions- FRQs

Torque AP Physics C Mechanics FRQ

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