Question

When a solid sphere is rolling along level surface the percentage of its total kinetic energy that is transnational is

• A. $29$
• B. $71$
• C. $60$
• D. $40$

Answer 1

The correct option is B $71$

we know,

$K{E}_{Total}=K{E}_{Translation}+K{E}_{rotation}$

$K{E}_{Translation}=\frac{M{V}^2}{2}$

$K{E}_{rotation}=\frac{1}{2}I{W}^2$

For solid sphere, $I=\frac{2}{5}M{R}^2$

For pure rolling, WR = v

So,

$K{E}_{Total}=\frac{m{v}^2}{2}+\left(\frac{1}{2}\right)\left(\frac{2}{5}\right)m(WR{)}^2$

$=\frac{m{v}^2}{2}+\frac{m{v}^2}{5}=\frac{7m{v}^2}{10}$

$\frac{K{E}_{translation}}{K{E}_{Total}}=\frac{1/2}{7/10}=\frac{m{v}^2}{m{v}^2}=$ $\frac{5\times 2}{14}=71\%$

option - B is correct.