AP Physics C Mechanics 5.5 Rotational Equilibrium Overview

[Maximum number: 8]

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 3 m hangs from the rod at Point P. The center of mass of the rod is located at Point C.

(a)

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 $F_{\mathrm{T}}$ in the horizontal string in terms of $L, m, \theta$ , and physical constants, as appropriate.

Figure for Question (a) — AP Physics C: Mechanics

Figure 2
Note: Figure not drawn to scale.

[ 3 ]

For indicating that the net torque exerted on the rod is equal to zero 1 point
Example Response

\Sigma \tau=0

For a correct expression for the torque exerted on the rod by the hanging mass 1 point
Example Response

3 m g \sin \theta\left(\frac{3 L}{8}\right)

For a correct expression for the torque exerted on the rod by the gravitational force 1 point
Example Response

m g \sin \theta\left(\frac{4 L}{8}\right)

For a correct expression for the torque exerted on the rod by the string 1 point
Example Responses

F_{\mathrm{T}} \sin \left(90^{\circ}-\theta\right)\left(\frac{6 L}{8}\right) \quad \text { OR } \quad F_{\mathrm{T}} \cos \theta\left(\frac{6 L}{8}\right)

Example Solution

\begin{aligned} & \Sigma \tau=0 \\ & 3 m g \sin \theta\left(\frac{3 L}{8}\right)+m g \sin \theta\left(\frac{4 L}{8}\right)-F_{\mathrm{T}} \sin \left(90^{\circ}-\theta\right)\left(\frac{6 L}{8}\right)=0 \\ & 3 m g \sin \theta\left(\frac{3 L}{8}\right)+m g \sin \theta\left(\frac{4 L}{8}\right)-F_{\mathrm{T}} \cos \theta\left(\frac{6 L}{8}\right)=0 \\ & \left(\frac{9}{8}\right) m g \sin \theta+\left(\frac{4}{8}\right) m g \sin \theta=\left(\frac{6}{8}\right) F_{\mathrm{T}} \cos \theta \\ & 13 m g \sin \theta=6 F_{\mathrm{T}} \cos \theta \\ & F_{\mathrm{T}}=\frac{13}{6} m g \tan \theta \end{aligned}

Scoring Note: A maximum of three points may be earned if the trigonometric functions ( $\sin$ and $\cos$ ) are reversed for all three torque terms.

(b)

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 $F_{\mathrm{T}, \text { new }}$ compare with the original tension $F_{\mathrm{T}}$ from part (b) ? Justify your reasoning.

Figure 3 Note: Figure not drawn to scale.

[ 2 ]

For indicating that the torque exerted on the rod by the string is always the same

1 point

For stating that as the angle between the string and the rod increases, the force exerted on the rod by the string decreases

1 point

Example Response

Because the torque exerted on the rod by the string is always the same, as the angle between the string and the rod increases, the tension $F_{T, \text { new }}$ must decrease.

Total for part (c)

2 points

(c)

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 m and a linear mass density $\lambda(x)=A+B x$ , where x is the distance from the pivot, $A=6.0 \mathrm{~kg} / \mathrm{m}$ , and $B=10.0 \mathrm{~kg} / \mathrm{m}^{2}$ .

[ 3 ]

(i)

Calculate the mass of the rod.

[ 3 ]

For indicating the total mass is the sum of differentiable masses along the length of the rod

1 point

Example Response
$M=\int d m$

For correctly writing d m in terms of x

1 point

Example Response
$M=\int_{0}^{1.2}(6+10 x) d x$

For a correct numeric answer with correct units

1 point

Example Response
$M=14.4 \mathrm{~kg}$

Example Solution
$\begin{aligned} M=\int d m M=\int \lambda d x M=\int_{0}^{1.2}(6+10 x) d x M=\left.\left(6 x+\frac{10 x^{2}}{2}\right)\right|_{0} ^{1.2 \mathrm{~m}} M=6 \mathrm{~kg} / \mathrm{m}(1.2 \mathrm{~m})+\frac{10 \mathrm{~kg} / \mathrm{m}^{2}(1.2 \mathrm{~m})^{2}}{2} M=14.4 \mathrm{~kg} \end{aligned}$

AP Physics C: Mechanics 5.5 Rotational Equilibrium and Newtons First Law in Rotational Form Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (i)