Rolling Motion of Cylinders with Different Mass Distributions

To determine the correct statement(s), let us analyze the rotational motion of the two solid cylinders $P$ and $Q$ rolling down an inclined plane without slipping.

1. Comparison of Moments of Inertia

Let both cylinders have mass $M$ and radius $R$.

• Cylinder $P$ has most of its mass concentrated near its outer surface, meaning its mass is distributed at larger distances from the axis of rotation. Thus, its moment of inertia about the central axis, $I_P$, is larger.
• Cylinder $Q$ has most of its mass concentrated near its central axis, meaning its mass is distributed closer to the axis of rotation. Thus, its moment of inertia, $I_Q$, is smaller.

Therefore, we have: $I_P > I_Q$

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2. Linear Acceleration

Consider a cylinder of mass $M$, radius $R$, and moment of inertia $I$ rolling down a fixed inclined plane of angle $\theta$.

Using Newton's second law for translational and rotational motion: $M g \sin\theta - f = M a$ $\tau = f R = I \alpha$

Since the cylinder rolls without slipping, the condition for pure rolling is $a = \alpha R$. Substituting $\alpha = \frac{a}{R}$ into the torque equation gives: $f = \frac{I a}{R^2}$

Substituting the friction force $f$ back into the translational equation: $M g \sin\theta - \frac{I a}{R^2} = M a$ $a = \frac{g \sin\theta}{1 + \frac{I}{M R^2}}$

Since $I_P > I_Q$, the denominators satisfy: $\left(1 + \frac{I_P}{M R^2}\right) > \left(1 + \frac{I_Q}{M R^2}\right)$

Hence, the linear accelerations are related by: $a_P < a_Q$

• Analysis of Option A: The time $t$ taken to travel a length $S$ along the incline starting from rest is $t = \sqrt{\frac{2S}{a}}$. Since $a_Q > a_P$, cylinder $Q$ reaches the bottom in less time than cylinder $P$ ($t_Q < t_P$). Thus, Option A is incorrect.
• Analysis of Option B: Cylinder $Q$ has a greater linear acceleration than cylinder $P$ ($a_Q > a_P$). Thus, Option B is incorrect.

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3. Linear Velocity and Translational Kinetic Energy

The linear velocity $v$ at the bottom of an incline of height $h$ (where $S = \frac{h}{\sin\theta}$) is given by: $v^2 = 2 a S = \frac{2 g h}{1 + \frac{I}{M R^2}}$

Since $I_P > I_Q$, we find that: $v_Q > v_P$

The translational kinetic energy at the bottom is given by: $K_{\text{trans}} = \frac{1}{2} M v^2$

Since $v_Q > v_P$ and both cylinders have equal mass $M$: $K_{\text{trans}, Q} > K_{\text{trans}, P}$

• Analysis of Option C: Cylinder $Q$ reaches the bottom with greater translational kinetic energy than cylinder $P$. Thus, Option C is incorrect.

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4. Angular Speed

For pure rolling, the relationship between linear speed $v$ and angular speed $\omega$ is: $\omega = \frac{v}{R}$

Since $v_Q > v_P$ and both cylinders have equal radius $R$: $\omega_Q = \frac{v_Q}{R} > \frac{v_P}{R} = \omega_P$

• Analysis of Option D: Cylinder $Q$ reaches the bottom with a greater angular speed ($\omega_Q > \omega_P$). Thus, Option D is correct.

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Conclusion

The only correct option is D.