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Q.
A sphere rolls down an inclined plane without slipping. What fraction of its total energy is rotational ?
Bihar CECEBihar CECE 2009System of Particles and Rotational Motion
Solution:
When a body rolls down an inclined plane, it makes rotational as well as translational motions. Thus, it is associated with rotational and translational kinetic energies. Hence, total kinetic energy of sphere
$K =K_{\text {rot. }}+K_{\text {trans }}$
$=\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}$
$=\frac{1}{2} \frac{2}{5} m r^{2}\left(\frac{v^{2}}{r^{2}}\right)+\frac{1}{2} m v^{2}$
$=\frac{1}{5} m v^{2}+\frac{1}{2} m v^{2}=\frac{7}{10} m v^{2}$
Also, $K_{\text {rot. }}=\frac{1}{2} I \omega^{2}=\frac{1}{5} m v^{2}$
Hence, $\frac{K_{\text {rot. }}}{K}=\frac{\frac{1}{5} m v^{2}}{\frac{7}{10} m v^{2}}=\frac{2}{7}$