AP Physics C Mechanics - 5.2 Connecting Linear and Rotational Motion- Exam Style questions- FRQs
Connecting Linear and Rotational Motion AP Physics C Mechanics FRQ
Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
Question
ii. Using the straight line, calculate the rotational inertia of the unknown object \(I_{U}\) about a vertical axis passing through its center of mass.
Most-appropriate topic codes (AP Physics C: Mechanics):
• Topic \(5.2\) — Connecting Linear and Rotational Motion (Part \( \mathrm{c} \))
• Topic \(5.6\) — Newton’s Second Law in Rotational Form (Part \( \mathrm{a} \))
• Topic \(6.1\) — Rotational Kinetic Energy (Parts \( \mathrm{b} \), \( \mathrm{f} \))
• Topic \(6.4\) — Conservation of Angular Momentum (Parts \( \mathrm{d} \), \( \mathrm{e} \), \( \mathrm{g} \))
▶️ Answer/Explanation
(a)
Using the rotational form of Newton’s second law, we set the net torque equal to the product of force and distance \(FD\). Then, substituting the derived angular acceleration into the rotational kinematics equation gives the angular speed \(\omega_P\).
\(\tau=I_{P}\alpha\)
\(FD=I_{P}\alpha \implies \alpha=\dfrac{FD}{I_{P}}\)
\(\omega_{P}=\omega_{0}+\alpha\Delta t\)
\(\omega_{P}=0+\left(\dfrac{FD}{I_{P}}\right)\Delta t\)
\(\boxed{\omega_{P}=\dfrac{FD\Delta t}{I_{P}}}\)
(b)
The kinetic energy of the rotating platform is given by the standard equation for rotational kinetic energy. We substitute the angular speed derived from part (a) and simplify to get the expression in terms of the given variables.
\(K=\dfrac{1}{2}I\omega_{P}^{2}\)
\(K=\dfrac{1}{2}I\left(\dfrac{FD\Delta t}{I}\right)^{2}\)
\(\boxed{K=\dfrac{(FD\Delta t)^{2}}{2I}}\)
(c)
Since the wheel drives the platform without slipping, their linear speeds at the point of contact must be exactly equal. We equate the linear speed of the wheel to the linear speed of the platform at distance \(D\) and then isolate \(\omega_{W}\).
\(v_{P}=v_{W}\)
\(D\omega_{P}=r\omega_{W}\)
\(\boxed{\omega_{W}=\dfrac{D\omega_{P}}{r}}\)
(d)
The collision between the falling disk and the platform is an inelastic rotational collision where angular momentum is strictly conserved. Equating the initial angular momentum to the final angular momentum of the combined system yields the final angular speed \(\omega_f\).
\(L_{i}=L_{f}\)
\(I_{P}\omega_{P}=(I_{P}+I_{P})\omega_{f}\)
\(I_{P}\omega_{P}=(2I_{P})\omega_{f}\)
\(\boxed{\omega_{f}=\dfrac{1}{2}\omega_{P}}\)
(e)(i)
Draw a straight line through the data points on the provided graph that best represents the overall trend. You should aim to keep roughly an equal number of points above and below the line.
(e)(ii)
By applying the conservation of angular momentum, we can express \(I_{U}\) in terms of the initial and final angular speeds. Taking two coordinate points from your drawn best-fit line to estimate the
ratio, we calculate the rotational inertia of the unknown object.
\(L_{i}=L_{f} \implies I_{P}\omega_{i}=(I_{P}+I_{U})\omega_{f}\)
\(I_{U}=I_{P}\left(\dfrac{\omega_{i}}{\omega_{f}}\right)-I_{P}\)
\(I_{U}=(3.1\,\text{kg}\cdot\text{m}^{2})\left(\dfrac{9.3\,\text{rad/s}-2.4\,\text{rad/s}}{4.0\,\text{rad/s}-1.0\,\text{rad/s}}\right)-(3.1\,\text{kg}\cdot\text{m}^{2})\)
\(\boxed{I_{U} \approx 4.1\,\text{kg}\cdot\text{m}^{2}}\)
(f)
Correct choice:
\(\boxed{K_{f}<K_{i}}\)
When the unknown object drops onto the platform, they eventually rotate together at the exact same angular velocity. This interaction makes it a perfectly inelastic rotational collision, and in any completely inelastic collision, some macroscopic kinetic energy is inherently dissipated.
(g)
Correct choice:
\(\boxed{\text{Greater than}}\)
The experimental value obtained is greater than the actual value about its center of mass. Since the object’s center of mass is misaligned with the axis of rotation, by the parallel axis theorem (\(I = I_{CM} + Mh^2\)), its effective rotational inertia about the platform’s axis is greater than its true rotational inertia about its own center of mass.