Rotational Motion: Conceptual Doubts Cleared (4)

Question

A solid sphere and a hollow sphere of equal mass and radius are released from rest at the top of the same incline. Which one reaches the bottom first, and why? Also, does the answer depend on the mass or the radius?

Solution — Step by Step

Final answer: solid sphere wins; result is independent of $m$ and $R$.

Why This Works

Rolling-without-slipping converts gravitational PE into both translational KE ($\tfrac{1}{2}mv^2$) and rotational KE ($\tfrac{1}{2}I\omega^2$). With $\omega = v/R$, the split depends only on $I/(mR^2)$, which is a pure number for a given shape. Mass and radius cancel out.

This is why competitive exams love this problem — the “simple” version is mass-independent, but examiners flip it by adding friction, slipping, or different inclines.

Alternative Method

Use energy conservation. Drop the sphere through height $h$. Then $mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2$. Substitute $\omega = v/R$ and solve for $v^2$:

$v^2 = \frac{2gh}{1 + I/(mR^2)}$

Larger $v^2$ means greater speed at the bottom — the solid sphere wins. Same conclusion, no Newton’s laws needed.

Common Mistake

Some students write $a = g\sin\theta$ and ignore the rotational term entirely. That formula is for sliding, not rolling. The presence of static friction (which provides torque without doing work in pure rolling) is what changes the answer.