Question

A solid sphere of mass $1kg$ and radius $10cm$ rolls down an inclined plane of height $7m$. The velocity of its centre as it reaches the ground level is

• A. $7m{s}^{-1}$
• B. $10m{s}^{-1}$
• C. $15m{s}^{-1}$
• D. $20m{s}^{-1}$

Answer 1

Answer: (B)

When a body of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ and angle of inclination $\theta$, it loses potential energy. However, it acquired both linear and angular speeds.
$mgh=\frac{1}{2}m{v}_{CM}^2+\frac{1}{2}I{\omega }^2$
For pure rolling $\omega =\frac{v_{CM}}{R}$ and using $I=m{k}^2$,
$mgh=\frac{1}{2}m{v}_{CM}^2+\frac{1}{2}m{k}^2\frac{v_{CM}^2}{R^2}$
Velocity at the lowest point, $v_{CM}=\sqrt{\frac{2gh}{1+\left(k^2/R^2\right)}}$
For solid sphere, $M{k}^2=\frac{2}{5}M{R}^2$
$\Rightarrow \frac{k^2}{R^2}=\frac{2}{5}$
$\therefore v=\sqrt{\frac{2\times 10\times 7}{1+(2/5)}}=10m{s}^{-1}$