Rotational inertia $$I$$ is directly proportional to mass $$M$$ (e.g., $$I=kMR^2$$). If the mass is doubled while shape and dimensions are unchanged, the new rotational inertia $$I'$$ will be $$2I$$. From Newton's second law for rotation, $$\alpha = \tau/I$$. The new angular acceleration $$\alpha'$$ will be $$\alpha' = \tau/I' = \tau/(2I) = \frac{1}{2}(\tau/I) = \frac{1}{2}\alpha$$.

According to Newton's second law for rotation, $$\alpha = \tau_{net}/I$$. Since both objects experience the same net torque, the object with the smaller rotational inertia will have the greater angular acceleration. The rotational inertia of a solid disk is $$I_{disk} = \frac{1}{2}MR^2$$, while that of a hoop is $$I_{hoop} = MR^2$$. Thus, $$I_{disk} < I_{hoop}$$, which means $$\alpha_{disk} > \alpha_{hoop}$$.

Newton's second law for rotation states that the net torque about an axis equals the rotational inertia about that axis times the angular acceleration. When considering torques about the center of mass, the applied force $$F$$ passes through the axis, so its lever arm is zero and it produces no torque. The only force that produces a torque about the center of mass is the static friction force $$f$$, which acts at the point of contact, a distance $$R$$ from the center. Thus, the net torque is $$\tau_{net} = fR$$, and the correct equation is $$fR = I\alpha$$.

The torque is provided by the component of the gravitational force perpendicular to the rod, acting at the center of mass ($$L/2$$ from the pivot). The torque is $$\tau = (Mg\sin\theta)(L/2)$$. Using $$\tau = I\alpha$$, we have $$\frac{1}{2}MgL\sin\theta = (\frac{1}{3}ML^2)\alpha$$. Solving for $$\alpha$$ gives $$\alpha = \frac{MgL\sin\theta/2}{ML^2/3} = \frac{3g\sin\theta}{2L}$$.

The linear equation of motion is $$Mg\sin\theta - f = Ma$$, where $$f$$ is the static friction force. The torque equation about the center of mass is $$\tau = fR = I\alpha$$. For rolling without slipping, $$a = R\alpha$$. Substituting $$f = I\alpha/R = Ia/R^2$$ into the linear equation gives $$Mg\sin\theta - \frac{Ia}{R^2} = Ma$$. Solving for $$a$$ yields $$a = \frac{Mg\sin\theta}{M + I/R^2}$$. With $$I = \frac{1}{2}MR^2$$, this becomes $$a = \frac{g\sin\theta}{1 + 1/2} = \frac{2}{3}g\sin\theta$$. Finally, $$\alpha = a/R = \frac{2g \sin\theta}{3R}$$.

This question assesses the cause of the change in angular velocity from a force/torque perspective. For the person to increase their angular velocity to match the turntable's rotation as they walk outward, the turntable must exert a torque on them in the direction of rotation. By Newton's third law, the person exerts an equal and opposite torque on the turntable. This torque opposes the turntable's rotation, causing a negative angular acceleration ($$\alpha = \tau_{net}/I$$), which means the turntable's angular velocity decreases. While angular momentum is conserved for the system, the question asks about the cause of the change in the turntable's velocity in terms of torque.

Newton's second law for rotation states that the net torque is equal to the rotational inertia times the angular acceleration, $$\tau_{net} = I\alpha$$. The angular acceleration is the second time derivative of the angular position, $$\alpha = \frac{d^2\theta}{dt^2}$$. The net torque is the restoring torque from the wire, $$\tau_{net} = -\kappa\theta$$. Substituting these into the rotational dynamics equation gives $$-\kappa\theta = I\frac{d^2\theta}{dt^2}$$. Rearranging this equation yields $$I\frac{d^2\theta}{dt^2} + \kappa\theta = 0$$, which is the standard form for the differential equation of simple harmonic motion.

For the putty to be accelerated from zero angular velocity to the final angular velocity of the disk, it must experience a net torque. This torque is provided by the contact force from the disk. The component of this force that is tangential to the circular path of the putty creates the torque that changes the putty's angular momentum. By Newton's third law, the putty exerts an equal and opposite tangential force on the disk, creating a torque that slows the disk down. The vertical forces (gravity, normal force) do not exert torques about the vertical axis. The radial force provides centripetal acceleration but no torque about the center.

This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. Newton's Second Law in rotational form is expressed as τ = Iα, indicating that net torque causes angular acceleration proportional to the moment of inertia. When the moment of inertia I is constant, the relationship shows that torque and angular acceleration are directly proportional. Choice B is correct because increasing τ_net while keeping I constant will proportionally increase α, as α = τ_net/I demonstrates a direct linear relationship. Choice A is incorrect because it suggests an inverse relationship, contradicting the fundamental equation. To help students: Draw parallels to F = ma in linear motion; use numerical examples showing how doubling torque doubles angular acceleration; emphasize that moment of inertia plays the same role as mass in resisting changes to rotational motion.

This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. Torque (τ) is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the lever arm and θ is the angle between force and lever arm. For a tangential force (θ = 90°), the torque equation simplifies to τ = rF since sin90° = 1. Choice A is correct because when r doubles while F remains constant and tangential, the torque τ = rF also doubles, demonstrating the direct proportionality between torque and lever arm. Choice B is incorrect because it suggests an inverse relationship, which would only occur if we were considering angular acceleration with changing moment of inertia. To help students: Use visual demonstrations showing how increasing the lever arm increases the rotational effect; emphasize that torque is directly proportional to both force and lever arm distance; practice problems where students identify which variables change and predict the effect on torque.