Question

A solid sphere of mass $M$ and radius $2R$ rolls down an inclined plane of height $h$ without slipping. The speed of its center of mass when it reaches the bottom is

• A. $\sqrt{\frac{6}{7}gh}$
• B. $\sqrt{3gh}$
• C. $\sqrt{\frac{10}{7}gh}$
• D. $\sqrt{\frac{4}{3}gh}$

Answer 1

Answer: (C)

$\sqrt{\frac{10}{7}gh}$

When solid sphere rolls down on an inclined plane, then it has both rotational and transitional kinetic energy
$K=K_{rot}+K_{trans}$
or $K=\frac{1}{2}I{\omega }^2+\frac{1}{2}M{v}^2$
where, $l=$ moment of inertia of solid sphere $=\frac{2}{5}M{R}^2$
$\therefore K=\frac{1}{2}\left(\frac{2}{5}M{\left(2R\right)}^2\right){\omega }^2+\frac{1}{2}M{v}^2$ $[\because R=2R]$
$=\frac{4}{5}M{R}^2{\left(\frac{V}{2R}\right)}^2+\frac{7}{2}M{v}^2$ $[\because v=r\omega ]$
$=\frac{1}{5}M{v}^2+\frac{1}{2}M{v}^2=\frac{7}{10}M{v}^2$
Now, gain in KE $=$ loss in PE
$\frac{7}{10}M{v}^2=Mgh$
$v=\sqrt{\frac{10}{7}gh}$