Rotational Motion: Common Mistakes and Fixes (5)

Question

A solid sphere of mass $M$ and radius $R$ rolls without slipping down a rough incline of angle $\theta$. A student writes the linear acceleration as $a = g\sin\theta$ and gets the answer wrong. What is the correct acceleration, and where did the student slip up?

Solution — Step by Step

Why This Works

When a body rolls without slipping, part of the available potential energy goes into rotational kinetic energy, so the centre-of-mass acceleration is less than for pure sliding. The factor $\frac{1}{1 + I/MR^2}$ encodes how much of the energy “leaks” into rotation. For a solid sphere, $I = \frac{2}{5}MR^2$, giving the famous $5/7$ factor.

For a solid cylinder, the factor is $2/3$. For a hollow shell, $3/5$. PYQ favourite — memorise these.

Alternative Method

Use energy conservation. After dropping height $h$, the centre has speed $v$:

$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}Mv^2$

So $v^2 = \frac{10}{7}gh$. Using $v^2 = 2as$ where $s = h/\sin\theta$, we recover $a = \frac{5}{7}g\sin\theta$.

Common Mistake

Students confuse $I$ about the centre of mass with $I$ about the contact point. For torque about the centre, use $I_{\text{cm}} = \frac{2}{5}MR^2$. If we instead take torque about the instantaneous contact point, only gravity’s torque matters and $I_{\text{contact}} = \frac{7}{5}MR^2$ — both routes give the same $a$, but mixing them up gives nonsense.