Question

A solid sphere mass $M$ and radius $R$ rolls on a horizontal surface without slipping. The ratio of rotational $K.E.$ ro total $K.E.$ is:-

• A. $\frac{1}{2}$
• B. $\frac{3}{7}$
• C. $\frac{2}{7}$
• D. $\frac{2}{10}$

Answer 1

The correct option is C $\frac{2}{7}$

Angular momentum energy

mass of sphere = m

radius = r

M.O.I $=\frac{2}{5}m{r}^2$

total energy $=K_R+K_T$

$K_{Total}=\frac{1}{2}I{w}^2+\frac{1}{2}m{v}^2$

$\downarrow$

As we know

$\frac{1}{2}\times \frac{2}{5}m{r}^2\left(\frac{v^2}{r^2}\right)+\frac{1}{2}m{v}^2$

$V=rw$

$K_{total}=\frac{1}{2}\left(\frac{7}{5}\right)m{v}^2$

Fraction of K.E of rotation

$=\frac{K_R}{K_{total}}$

$=\frac{\frac{1}{2}\left(\frac{2}{5}\right)M{V}^2}{\frac{1}{2}\left(\frac{7}{5}\right)M{V}^2}=\frac{2}{7}$