Question

A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is

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

Answer 1

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

${KE}_{Total}$ = $\frac{1}{2}I{W}^2+\frac{1}{2}M{V}^2$

$=\frac{1}{2}\times \frac{2}{5}M{R}^2{W}^2+\frac{1}{2}M{V}^2$

$=\frac{1}{5}M{R}^2{W}^2+\frac{1}{2}M{V}^2$ =$\frac{1}{5}M{V}^2+\frac{1}{2}M{V}^2$

$=\frac{7}{10}M{V}^2$

${KE}_{rot}=\frac{1}{2}I{W}^2=\frac{1}{5}M{V}^2$

$\frac{{KE}_{rot}}{{KE}_{Total}}=\frac{\frac{1}{5}M{V}^2}{\frac{7}{10}M{V}^2}$

$=\frac{2}{7}$.