Question

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

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

Answer 1

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

Given, Mass, $m=500g$

Speed, $v=5.00cm/sec$

As we know, kinetic energy in pure rolling is,
$KE=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Therefore, putting all value as mention in diagram in step $1$.
Taking $\frac{1}{2}$ as common from $K.E$ equation, we get
$KE=\frac{1}{2}(\frac{2}{5}m{R}^2+m{R}^2){\left(\frac{v}{R}\right)}^2$

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

Hence,
$KE=\frac{35}{4}\times {10}^{-4}$ Joule

Or $KE=8.75\times {10}^{-4}$ Joule