Animated Solution for Physics - Rotational Motion: Two solid cylinders $P$ and $Q$ of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder $P$ has most of its mass concentrated near its surface, while $Q$ has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

A Tale of Two Cylinders

Imagine two cylinders, $P$ and $Q$, poised at the top of an inclined plane, ready for a race to the bottom. They look identical from the outside—they share the exact same mass $m$ and the exact same radius $R$. However, their internal architecture tells a different story.

Cylinder $P$ has its mass pushed out towards its surface, making it behave dynamically like a hollow pipe or a ring. On the other hand, cylinder $Q$ has its mass packed tightly near its central axis, acting more like a solid log or a disc. This subtle difference in mass distribution is the key to unlocking the entire problem.

Acceleration in Pure Rolling

To determine who wins the race, we must look at their resistance to rotational motion, known as the moment of inertia ($I$). The moment of inertia is defined by the integral $I=\int r^{2}dm$. Because the mass of cylinder $P$ is located at a greater average distance $r$ from the axis of rotation, it possesses a significantly higher moment of inertia than cylinder $Q$. Mathematically, we can state with certainty that $I_P>I_Q$.

Now, let's bring in the master equation for a body rolling down an incline without slipping. The linear acceleration $a$ of the center of mass is given by:

Look closely at the denominator. The term $\frac{I}{m{R}^2}$ acts as a "sluggishness factor." Since $I_P$ is larger, the denominator for cylinder $P$ is larger, which inevitably makes its overall linear acceleration smaller. Therefore, we conclude that $a_Q>a_P$.

Velocity and Angular Speed

Because cylinder $Q$ accelerates at a greater rate, it will undoubtedly win the race, reaching the bottom of the incline in less time than cylinder $P$. Consequently, as it crosses the finish line, its linear velocity will be higher: $v_Q>v_P$.

But the question specifically asks about their angular speeds. In a state of pure rolling, the linear velocity $v$ and the angular speed $\omega$ are locked in a strict relationship: $v=\omega R$. Rearranging this gives us $\omega =\frac{v}{R}$.

Since both cylinders share the identical radius $R$, the one with the higher linear velocity must also possess the higher angular speed. Thus, ${\omega }_Q>{\omega }_P$. Cylinder $Q$ not only wins the race linearly but is also spinning more furiously at the bottom.

It is also fascinating to view this race through the lens of energy conservation. Both cylinders start from the same height, meaning they possess the exact same initial gravitational potential energy ($mgh$). Assuming no energy is lost to non-conservative forces (static friction does no work in pure rolling), they must both arrive at the bottom with the exact same total kinetic energy.

However, this total kinetic energy is split into two bank accounts: translational kinetic energy ($\frac{1}{2}m{v}^2$) and rotational kinetic energy ($\frac{1}{2}I{\omega }^2$). Because cylinder $Q$ is moving faster linearly ($v_Q>v_P$), its translational kinetic energy is greater. To balance the books, cylinder $P$ must have a greater rotational kinetic energy, compensating for its slower linear pace.

Ultimately, statement (d) is the only correct conclusion: Cylinder $Q$ reaches the ground with a larger angular speed.