Question

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

• A. $7\, ms ^{-1}$
• B. $10\, ms ^{-1}$
• C. $15\, ms ^{-1}$
• D. $20\, ms ^{-1}$

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.
$m g h=\frac{1}{2} m v_{ CM }^{2}+\frac{1}{2} I \omega^{2}$
For pure rolling $\omega=\frac{v_{ CM }}{R}$ and using $I=m k^{2}$,
$m g h=\frac{1}{2} m v_{ CM }^{2}+\frac{1}{2} m k^{2} \frac{v_{ CM }^{2}}{R^{2}}$
Velocity at the lowest point, $v_{ CM }=\sqrt{\frac{2 g h}{1+\left(k^{2} / R^{2}\right)}}$
For solid sphere, $M k^{2}=\frac{2}{5} M R^{2}$
$\Rightarrow \frac{k^{2}}{R^{2}}=\frac{2}{5}$
$\therefore v=\sqrt{\frac{2 \times 10 \times 7}{1+(2 / 5)}}=10\, ms ^{-1}$