Question

A solid sphere of mass $1kg$ rolls without slipping on a plane surface. Its kinetic energy is $7\times 10^{-3}J$. The speed of the centre of mass of the sphere is ___$cm{s}^{-1}$

Answer 1

Answer: 10

$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=7\times 10^{-3}$
$\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{V}{R}\right)}^2=7\times 10^{-3}$
$\frac{1}{2}M{V}^2\left[1+\frac{2}{5}\right]=7\times 10^{-3}$
$\frac{1}{2}(1)\left(V^2\right)\left(\frac{7}{5}\right)=7\times 10^{-3}$
$V^2=10^{-2}$
$V=10^{-1}=0.1m/s=10cm/s$