Question

A uniform sphere of mass $500\text{g}$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00\text{cm/s}$. Its kinetic energy is

• A. $8.75\times {10}^{-4}\text{J}$
• B. $8.75\times {10}^{-3}\text{J}$
• C. $6.25\times {10}^{-4}\text{J}$
• D. $1.13\times {10}^{-3}\text{J}$

Answer 1

The correct option is A $8.75\times {10}^{-4}\text{J}$

$K.E$ of the sphere = translational $K.E+\text{rotational K.E}$

$=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$\text{Where, I = moment of inertia,}$
$\omega =\text{Angular, velocity of rotation}$
$\text{m = mass of the sphere}$
$\text{v = linear velocity of centre of mass of sphere}$

$\because \text{Moment of intertia of sphere}I=\frac{2}{5}m{R}^2$

$\therefore K.E=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{R}^2\times {\omega }^2$

$\Rightarrow K.E=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{R}^2\times {\left(\frac{v}{R}\right)}^2(\because \omega =\frac{v}{R})$

$\Rightarrow KE=\frac{1}{2}(\frac{2}{5}m{R}^2+m{R}^2){\left(\frac{v}{R}\right)}^2$

$\Rightarrow KE=\frac{1}{2}m{R}^2\times \frac{7}{5}\times \frac{v^2}{R^2}=\frac{7}{10}\times \frac{1}{2}\times \frac{25}{{10}^4}$

$\Rightarrow KE=\frac{35}{4}\times {10}^{-4}\text{J}$

$\Rightarrow KE=8.75\times {10}^{-4}\text{J}$