Question

For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is \(\frac{x}{5}\). The value of x is____.

Answer 1

Correct Answer: 2

Solution:
For a rolling spherical shell, the angular velocity \( \omega = \frac{v}{R} \).
The rotational kinetic energy is given by: \[ K_{\text{rot}} = \frac{1}{2} \left( \frac{2}{3} m R^2 \right) \left( \frac{v}{R} \right)^2 = \frac{1}{2} \left( \frac{2}{3} m R^2 \right) \left( \frac{v^2}{R^2} \right) = \frac{1}{3} m v^2. \] The total kinetic energy is: \[ K_{\text{total}} = \frac{1}{2} mv^2 + \frac{1}{2} \left( \frac{2}{3} m R^2 \right) \left( \frac{v}{R} \right)^2 = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \frac{5}{6} mv^2. \] The ratio of rotational kinetic energy to total kinetic energy is: \[ \frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3} m v^2}{\frac{5}{6} m v^2} = \frac{2}{5}. \] Thus, \( \frac{x}{5} = \frac{2}{5} \), so \( x = 2 \).