Answer: (A)

Connecting conservation of angular momentum with changes in rotational inertia reveals the system's behavior. Since no external torques act on the star, L remains constant. Rotational inertia I is proportional to R^2, so halving the radius reduces I to (1/2)^2 = 1/4 of its initial value. To conserve L (where L = Iw), w must increase by a factor of 4.

Answer: (A)

Analyzing the impulse transfer and system boundaries distinguishes the two outcomes. The bouncing ball delivers a greater impulse (change in momentum from +p to -p) than the sticking clay (+p to 0), causing a larger change in the rod's angular momentum. However, for the system (rod + object), net external torque is zero, so total angular momentum is conserved in both trials.

Answer: (A)

Combining orbital mechanics with definitions of L and K reveals the scaling relationships. Orbital speed v scales with 1/sqrt(r), so v_B is half of v_A. Angular momentum L = mvr scales with sqrt(r), making L_B double L_A. Kinetic energy K scales with v^2 (and thus 1/r), making K_B one-fourth of K_A.

Answer: (A)

Evaluating how mass distribution affects rotational inertia determines the comparison. A hollow sphere has mass concentrated further from the axis than a solid sphere, giving it a larger rotational inertia I. Since L = Iw and K = 0.5Iw^2, and w is identical for both, the hollow sphere has greater L and K.

Answer: (A)

Angular momentum is $L = mvr$. Since $v \propto 1/\sqrt{r}$, $L \propto r(1/\sqrt{r}) = \sqrt{r}$. Increasing $r$ by a factor of 4 increases $L$ by a factor of $\sqrt{4} = 2$.

Answer: (A)

Kinetic energy is $K = \frac{1}{2}mv^2$. Since $v = \sqrt{GM/R}$, $v^2 \propto 1/R$, and thus $K \propto 1/R$. Doubling the radius halves the kinetic energy.

Answer: (A)

Rotational kinetic energy depends on the square of the angular velocity ($K = \frac{1}{2}I\omega^2$). Tripling the angular velocity results in a factor of $3^2 = 9$ increase in energy.

Answer: (A)

Rotational inertia for a sphere is proportional to the square of the radius ($I \propto R^2$). Doubling the radius quadruples $I$, which linearly quadruples the kinetic energy ($K \propto I$).

Answer: (A)

Rotational inertia decreases by a factor of 4 ($I \propto R^2$). Since angular momentum $L$ is conserved, $K = \frac{L^2}{2I}$ increases by a factor of 4 (inverse relationship with $I$).

Answer: (A)

Rotational kinetic energy for a rolling object is $K = \frac{1}{2}I\omega^2 = \frac{1}{2}(cMR^2)(v/R)^2 = \frac{1}{2}cMv^2$. The radius $R$ cancels out, so energy depends only on mass and linear speed.