Question

A hollow cylinder and a solid cylinder of same mass and same external radius are rolling without slipping down an inclined plane of inclination $\theta$ from horizontal. If both of them are released at the same time, then which one reaches the bottom first?

• A. Solid cylinder
• B. Hollow cylinder
• C. Both reaches simultaneously
• D. Can't say anything

Answer 1

The correct option is A Solid cylinder

For rolling without slipping, the linear acceleration $a$ of $COM$ of rolling body, down the inclined plane is given by:
$a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}....\left(i\right)$
where $I\to$moment of inertia of body about an axis passing through its $\text{COM}$

$\Rightarrow$Since, both have to cover same displacement down the inclined plane when they reaches bottom, time taken by each of them,
$t\propto \frac{1}{a}...\left(ii\right)$
From Eq. $\left(i\right)$,
$a_{\text{solid cylinder}}=\frac{g\sin \theta }{1+\frac{\left(\frac{M{R}^2}{2}\right)}{M{R}^2}}$
$\therefore a_{\text{solid cylinder}}=\frac{2g\sin \theta }{3}$

$a_{\text{Hollow cylinder}}=\frac{g\sin \theta }{1+\frac{\left(M{R}^2\right)}{M{R}^2}}$
$\therefore a_{\text{Hollow cylinder}}=\frac{g\sin \theta }{2}$
$\Rightarrow a_{\text{solid cylinder}}>a_{\text{Hollow cylinder}}$
From Eq. $\left(ii\right)$ we can infer that,
$t_{\text{solid cylinder}}<t_{\text{Hollow cylinder}}$
$\therefore$ Solid cylinder will reach the bottom first.