Q. A solid sphere is rolling down an inclined plane. Then, the ratio of its translational kinetic energy to its rotational kinetic energy is
J & K CETJ & K CET 2015System of Particles and Rotational Motion
Solution:
Translational kinetic energy,
$K_{T}=\frac{1}{2} m v^{2}$
Rotational kinetic energy $V_{1}$
$K_{R}=\frac{1}{2} I \omega^{2}=\frac{1}{2} \times \frac{2}{5} m R^{2} \times \frac{v^{2}}{R^{2}}$
$=\frac{1}{5} m v^{2}$
$\frac{K_{T}}{K_{R}}=\frac{\frac{1}{2} m v^{2}}{\frac{1}{5} m v^{2}}$
$=\frac{5}{2}=2.5$
