Rotational Motion: Step-by-Step Worked Examples (8)

Question

A solid sphere of mass $2 \, \text{kg}$ and radius $0.1 \, \text{m}$ rolls without slipping down an inclined plane of angle $30°$ and length $2 \, \text{m}$. Find its linear speed at the bottom. Take $g = 10 \, \text{m/s}^2$.

Solution — Step by Step

Why This Works

Energy conservation works here because friction does no work for pure rolling — the contact point is instantaneously at rest. So we never have to compute the friction force explicitly.

The factor $7/10$ for a solid sphere is worth memorising. Quick reference for the speed-at-bottom factor:

Alternative Method — Newton’s Laws

We could write the equation of motion along the incline ($mg\sin\theta - f = ma$) and the torque equation about the centre ($fR = I\alpha$), then use $a = R\alpha$ to eliminate $f$ and solve for $a$. Then $v^2 = 2aL$.

That gives $a = \frac{5g\sin\theta}{7} = \frac{5 \times 10 \times 0.5}{7} = \frac{25}{7} \, \text{m/s}^2$, and $v^2 = 2 \times \frac{25}{7} \times 2 = \frac{100}{7}$. Same answer.

Energy method is faster here. Use Newton when the question asks for friction or acceleration explicitly.

Common Mistake

Students forget the rotational kinetic energy and write $mgh = \frac{1}{2}mv^2$. This gives $v = \sqrt{2gh} \approx 4.47 \, \text{m/s}$ — the answer for sliding without rolling, not for rolling.