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Q.
A spherical shell rolls on a table without slipping. What is the percentage of its $K . E$ which is rotational?
System of Particles and Rotational Motion
Solution:
K.E of rotation $=\frac{1}{2} I \omega^{2}=\frac{1}{2} I \times \frac{v^{2}}{r^{2}}$
$=\frac{1}{2} \times \frac{2}{3} \times \frac{m \times r^{2} \times v^{2}}{r^{2}}$
K.E of rotation $=\frac{1}{3} m v^{2}$
Total energy $=\frac{1}{2} m v^{2}\left(1+\frac{K^{2}}{R^{2}}\right)$
$=\frac{1}{2} m v^{2}\left(1+\frac{2}{3}\right)$
$=\frac{5}{6} m v^{2} $
$\therefore \frac{\text { Rotational } K . E}{\text { Total Energy }} \times 100=\frac{\frac{1}{3} m v^{2}}{\frac{5}{6} m v^{2}} \times 100$
$=\frac{2}{5} \times 100$
$=40 \%$