Rotational Inertia AP Physics C Mechanics FRQ – Exam Style Questions etc.
Rotational Inertia AP Physics C Mechanics FRQ
Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
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 θ 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) 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, θ, 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 q 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 m and a linear mass density λ(x) = A + Bx.
where x is the distance from the pivot, $A = 6.0 kg/m$, and $B = 10.0 kg/m^{2}$.
i. Calculate the mass of the rod.
ii. Calculate the rotational inertia of the rod about the pivot.
▶️Answer/Explanation
Ans:-
(a) For drawing and appropriately labeling the downward forces that are exerted on the rod at points P and C
Scoring Note: Labeling the downward force of tension as $F_{block}$, 3 mg, or similar may earn this point.
Scoring Note: Examples of appropriate labels for the force due to gravity include: $F_{G}$ , $F_{g}$ ,
$F_{grav} $, W , mg , Mg,“grav force,” “ F Earth on block,” “ F on block by Earth,” $F_{Earth on block}$ ,
$F_{E,Block}$ . The labels G and g are not appropriate labels for the force due to gravity.
Scoring Note: Examples of appropriate labels for the force from the pivot include: $F_{p}$ ,
$F_{pivot}$, $F_{n}$, $F_{N }$, N , “normal force,” “pivot force.”
Scoring Note: Examples of appropriate labels for the tension force include $F_{t}$, $F_{T}$ , T ,
$F_{string}$, and “Force from string.”
(b) For indicating that the net torque exerted on the rod is equal to zero
Scoring Note: A maximum of three points may be earned if the trigonometric functions ( sin and cos ) are reversed for all three torque terms.
(c) For indicating that the torque exerted on the rod by the string is always the same
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 $F_{T, new}$ must decrease.
(d)(i) For indicating the total mass is the sum of differentiable masses along the length of the rod
Example Solution
$M=\int dm$
$M=\int \lambda dx$
$M=\int_{0}^{1.2}\left ( 6+10x \right )dx$
$M=\left ( 6+\frac{10x^{2}}{2} \right )\int_{0}^{1.2}$
$M=6kg/m(1.2m)+\frac{10kg/m^{2}(1.2m)^{2}}{2}$
$M=14.4kg$
(d)(ii) For a correct substitution of λ into an integral expression of rotational inertia
Question
A uniform disk of radius R and mass \(m_{d}\) is attached to a vertical pole by a horizontal axle that passes through the center of the disk. Friction between the axle and the disk is negligible. A lump of clay of mass \(m_{c}\) is attached to the edge of the disk at Point A. The size of the lump of clay is small compared with the radius of the disk. A horizontal string is connected from the pole to the edge of the disk at Point A. The string makes an angle \(\Theta\) with the line between Point A and the axle, as shown in Figure 1.
(a) On the following representation of the clay-disk system, draw and label the external forces (not components) exerted on the system. 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 system.
(b) Derive an expression for the tension \(F_{T}\) in the string when the clay is at Point A, as shown in Figure 1, in terms of R, \(m_{d}\), \(m_{c}\), \(\Theta\), and physical constants, as appropriate.
(c) The string remains connected to the edge of the disk at Point A. The clay is moved to Point B, which is horizontally in line with the axle, as shown in Figure 2. How does the new tension \(F_{T},_{new}\) compare with tension \(F_{T}\) from part (b)? Justify your reasoning.
(d) A nonuniform disk is now attached to the axle. The lump of clay is attached to the disk at Point B, as shown in Figure 3. The clay has mass \(m_{c}\) = 0.60 kg and the disk has a radius R = 0.30 m. The mass density of the disk varies radially and can be modeled by \(\rho \left ( r \right ) =\beta r\), where r is the radial distance from the axle and \(\beta \) = \(4.0kg/m^{3}\) .
i. Calculate the rotational inertia of the disk about the axle.
ii. The string connecting the disk to the pole is cut. Calculate the magnitude of the initial angular acceleration of the clay-disk system.
▶️Answer/Explanation
3(a)Example Responses
Scoring Note: Examples of appropriate labels for the gravitational force include \(F_{G}\), \(F_{g }\), \(F_{grav}\), W , mg , Mg , “grav force,” “ F Earth on disk,” “ F on disk by Earth,” \(F_{Earth\\\ on \\\ Disk}\), \(F_{E},_{Disk}\) , and \(F_{Disk},_{E}\) . The labels G or g are not appropriate labels for the gravitational force.
Scoring Note: Examples of appropriate labels for the normal force from the axle include \(F_{N}\), \(F_{axle }\), N , “normal force,” and “axle force.”
Scoring Note: Examples of appropriate labels for the tension force include \(F_{t}\), \(F_{T}\), T , \(F_{string}\), and “Force from string.”
3(b)Example Responses
Scoring Note: A maximum of three points can be earned if the trigonometric functions ( sin and cos ) are reversed for both torque expressions.
3(c)Example Responses
The torque caused by the weight of the clay at Point B is greater than when the clay is at Point A because the component of the weight that is perpendicular to the lever arm is larger. To maintain equilibrium, the net torque on the system is still zero, therefore the tension \(F_{T},_{new}\) must be greater.
3(d)(i)Example Solution
3(d)(ii)Example Solution
