Question

A uniform sphere of mass 500 g rolls without slipping on a plane surface so that its centre moves at a speed of 0.02 m/s. The total kinetic energy of rolling sphere would be (in J)

• A. $1.4\times {10}^{-4}J$
• B. $0.75\times {10}^{-3}J$
• C. $5.75\times {10}^{-3}J$
• D. $4.9\times {10}^{-5}J$

Answer 1

The correct option is A $1.4\times {10}^{-4}J$

The total kinetic energy of rolling sphere=Kinetic energy of translation+Kinetic energy of rotation

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

Since the sphere rolls without slipping,

$v=R\omega$

$⟹K=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\omega }^2$

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

$=\frac{7}{10}m{v}^2$

$=\frac{7}{10}\times 0.5\times {0.02}^{2}kg{m}^2/s^2$

$=1.4\times {10}^{-4}J$