Question

If a solid sphere of mass $1\text{kg}$ and radius $0.1\text{m}$ rolls without slipping at a uniform velocity of $1\text{m/s}$ along a straight line on a horizontal floor, then its kinetic energy is

• A. $\frac{7}{5}\text{J}$
• B. $\frac{2}{5}\text{J}$
• C. $\frac{7}{10}\text{J}$
• D. $1\text{J}$

Answer 1

The correct option is C $\frac{7}{10}\text{J}$

When a body is in state of pure rolling the condition which hold is $(v=r\omega )$ and it has translational $K.E$ as well as rotational $K.E$
$\therefore {K.E}_{total}={K.E}_{Trans}+{K.E}_{Rot}=\frac{1}{2}m{v}_{\text{CM}}^2+\frac{1}{2}{I}_{\text{CM}}{\omega }^2....\left(i\right)$
Where $v=v_{\text{CM}},\omega =\frac{v}{r}$ and $I_{\text{CM}}=\frac{2}{5}m{r}^2$ for solid sphere.
From Eq. $\left(i\right)$,
${K.E}_{total}=\frac{1}{2}m{v}^2+[\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)\times \frac{v^2}{r^2}]$
${K.E}_{total}=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2$
$\therefore {K.E}_{total}=\frac{7}{10}\times 1\times 1^2=\frac{7}{10}\text{J}$