Question

If a spherical balls rolls on a table without slipping, the fraction of its total energy associated with rotation is

• A. $\frac{3}{5}$
• B. $\frac{2}{7}$
• C. $\frac{2}{5}$
• D. $\frac{3}{7}$

Answer 1

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

If the spherical ball is not slipping then,
$V=\omega R$
The total KE $=$ Rotational KE $+$ Translational KE
$=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$
$=\frac{1}{2}\times \frac{2}{5}m{R}^2\times \frac{V^2}{R^2}+\frac{1}{2}m{v}^2$
Total KE $=\frac{1}{2}\left(\frac{2}{5}\right)m{V}^2+\frac{1}{2}m{V}^2=\frac{1}{2}m{V}^2\times \left(\frac{7}{5}\right)$
Rotational KE $=\frac{1}{2}m{V}^2\left(\frac{2}{5}\right)$
$\therefore$ Fraction $=\frac{RotationalKE}{TotalKE}=\frac{\frac{1}{2}m{V}^2\left(\frac{2}{5}\right)}{\frac{1}{2}m{V}^2\left(\frac{7}{5}\right)}=\frac{2}{7}$