Question

A spherical shell rolls on a table without slipping. What is the percentage of its $K.E$ which is rotational?

• A. $50\%$
• B. $40\%$
• C. $70\%$
• D. $90\%$

Answer 1

Answer: (B)

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 $=\frac{1}{2}m{v}^2\left(1+\frac{K^2}{R^2}\right)$
$=\frac{1}{2}m{v}^2\left(1+\frac{2}{3}\right)$
$=\frac{5}{6}m{v}^2$
$\therefore \frac{\text{ Rotational }K.E}{\text{ Total Energy }}\times 100=\frac{\frac{1}{3}m{v}^2}{\frac{5}{6}m{v}^2}\times 100$
$=\frac{2}{5}\times 100$
$=40\%$