Question

A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is \(\frac{x}{5}\). The value of x is __________.

Answer 1

Correct answer is : 2

\(\frac{\frac{1}{2} \mathrm{I} \omega^{2}}{\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{mv}^{2}}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^{2}\right) \omega^{2}}{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^{2}\right) \omega^{2}+\frac{1}{2} \mathrm{~m}(\mathrm{R} \omega)^{2}}\) 

\(=\frac{\frac{2}{3}}{\frac{2}{3}+1}=\frac{2}{5}\)

\(\mathrm{x}=2\)