Question

When a uniform solid sphere and a disc of the same mass and of the same radius rolls down without slipping on an inclined plane from rest to the same distance along the inclined plane, then the ratio of the time taken by solid sphere and the disc to cover the distance is

• A. $15:14$
• B. ${15}^2:{14}^2$
• C. $\sqrt{14}:\sqrt{15}$
• D. $14:15$

Answer 1

The correct option is C $\sqrt{14}:\sqrt{15}$

In case of pure rolling, the acceleration of $\text{COM}$ of the rolling bodies down the inclined plane is given by:
$a=\frac{g\sin \theta }{1+\frac{K^2}{r^2}}=\frac{g\sin \theta }{1+\frac{I}{m{r}^2}}....\left(i\right)$
Now, both bodies are starting from rest $u=0,{\omega }_i=0$, hence applying the kinematic equation for displacement $S$ covered along the inclined plane:
$S=\frac{1}{2}a{t}^2$
$\Rightarrow \frac{t_{\text{sphere}}}{t_{\text{disc}}}=\sqrt{\frac{a_{\text{disc}}}{a_{\text{sphere}}}}....\left(ii\right)$
From Eq. $\left(i\right)$ and $\left(ii\right)$:

$\frac{t_{sphere}}{t_{disc}}=\sqrt{\frac{(1+\frac{I_{\text{sphere}}}{m{r}^2})}{(1+\frac{I_{\text{disc}}}{m{r}^2})}}...\left(iii\right)$
Putting in Eq. $\left(iii\right)$ $I_{\text{disc}}=\frac{m{r}^2}{2},I_{\text{sphere}}=\frac{2}{5}m{r}^2$
$\Rightarrow \frac{t_{sphere}}{t_{disc}}=\sqrt{\frac{1+\frac{2}{5}}{1+\frac{1}{2}}}=\sqrt{\frac{14}{15}}$
$\therefore t_{sphere}:t_{disc}=\sqrt{14}:\sqrt{15}$