Question

A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

• A. Both together only when angle of inclination of the plane is ${45}^{\circ }$
• B. both together
• C. hollow cylinder
• D. solid cylinder

Answer 1

The correct option is D solid cylinder


For rolling without slipping (pure rolling), we have $a=\alpha R$ ... (1)

From, $\sum F=ma$
$mg\sin \theta -f=ma$ ... (2)
(Here $f$ is the friction force, $a=$ linear acceleration $\alpha =$ angular acceleration)

As we know,
$\tau =I\alpha$ (about COM)
$f\times R=I\left(\frac{a}{R}\right)$ [from eqn($1$)]
$f=\frac{I.a}{R^2}$ ... (3)

From eq. (2) and (3),
$mg\sin \theta -\frac{Ia}{R^2}=ma$
$\Rightarrow a=\frac{mg\sin \theta }{(\frac{I}{R^2}+m)}$
$\Rightarrow a=\frac{mg{R}^2\sin \theta }{I+m{R}^2}$
[$m,R$ and $\theta$ are constant]

Thus, $a\propto \frac{1}{I}$
i.e the body whose $I$ (MOI) is less will reach the bottom first.
As we know, $I_{\text{solid cylinder}}<I_{\text{hollow}}$
Hence, solid cylinder will reach the bottom first.