Question

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

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

Answer 1

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

When a body rolls over a smooth surface, it has linear K.E. and rotational K.E.
$\therefore E=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

where $\omega =\frac{v}{r}$ and $I=\frac{2}{5}m{r}^2$ for solid sphere.

$\therefore K.E.=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)\cdot \frac{v^2}{r^2}$

$=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2=\frac{7}{10}\times 1\times 1^2$

$=\frac{7}{10}J$