Rotational Motion: Application Problems (9)

Question

A solid sphere of mass $M = 2$ kg and radius $R = 0.1$ m rolls without slipping down an inclined plane of angle $\theta = 30°$. Find the linear acceleration of the centre of mass. Take $g = 10$ m/s$^2$.

This is a JEE Main favourite — appeared in 2022 Shift 2 and 2024 Shift 1 with different shapes. The approach is the same every time.

Solution — Step by Step

Why This Works

For a sliding block (no rotation), $a = g\sin\theta = 5$ m/s$^2$. For a rolling sphere, $a$ is reduced because part of the gravitational potential energy goes into rotational kinetic energy instead of translational.

The $\tfrac{5}{7}$ factor is a memorable signature of the solid sphere. Other shapes give:

• Hollow sphere: $a = \tfrac{3}{5}g\sin\theta$
• Solid cylinder: $a = \tfrac{2}{3}g\sin\theta$
• Hollow cylinder/ring: $a = \tfrac{1}{2}g\sin\theta$

Higher $I$ means more energy locked in rotation, so smaller linear $a$.

Alternative Method

Energy method. Potential energy lost = total kinetic energy gained:

$Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}Mv^2(1 + k)$

Differentiate w.r.t. time to get $a$, or use $v^2 = 2as$ with $h = s\sin\theta$ to confirm.

Common Mistake

Students forget to write the torque equation and just use $Mg\sin\theta = Ma$, getting $a = 5$ m/s$^2$. This ignores that rotation steals energy from translation. Always write both Newton’s second law AND the torque equation for rolling problems.