Question

A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ____ ms^–1.

Hint:

For rolling motion:
• Include both translational and rotational kinetic energies.
• Use the pure rolling condition ω = v/r to simplify calculations.

Answer 1

Correct Answer: 40

1. Kinetic Energy of a Rolling Sphere: - Total kinetic energy: \[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2, \] where \( I = \frac{2}{5} mr^2 \) for a sphere.
2. Relation Between \( v \) and \( \omega \): - For pure rolling: \[ \omega = \frac{v}{r}. \]
3. Substitute Values: \[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left( \frac{2}{5} mr^2 \right) \left( \frac{v^2}{r^2} \right). \] Simplify: \[ KE = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2. \]
4. Solve for \( v \): \[ 2240 = \frac{7}{10} \times 2 \times v^2, \] \[ v^2 = 1600, \quad v = 40 \, \text{ms}^{-1}. \]

Final Answer: 40 ms^-1