Question

A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameter with angular speed of 10 rad/s. If the moment of inertia of the sphere about its tangent is \( x \times 10^2 \) times its angular momentum about the diameter. Then the value of \( x \) will be_____

Hint:

Use the parallel axis theorem to calculate the moment of inertia about a tangent and then apply the ratio to find the desired value.

Answer 1

Correct Answer: 35

The moment of inertia of a solid sphere about its diameter is given by: \[ I = \frac{2}{5} m r^2 \] The moment of inertia of the sphere about its tangent (using parallel axis theorem) is: \[ I_t = \frac{2}{5} m r^2 + m r^2 = \frac{7}{5} m r^2 \] The angular momentum about the diameter is: \[ L_{\text{diameter}} = I \cdot \omega = \frac{2}{5} m r^2 \cdot \omega \] Now, the angular momentum about the tangent is: \[ L_{\text{tangent}} = I_t \cdot \omega = \frac{7}{5} m r^2 \cdot \omega \] The ratio of the angular momentum about the tangent to the diameter is: \[ \frac{L_{\text{tangent}}}{L_{\text{diameter}}} = \frac{\frac{7}{5} m r^2 \cdot \omega}{\frac{2}{5} m r^2 \cdot \omega} = \frac{7}{2} \] This is given as \( x \times 10^2 \). Thus: \[ x \times 10^2 = \frac{7}{2} \] Solving for \( x \): \[ x = \frac{7}{2} \times 10^2 = 35 \] Thus, the value of \( x \) is 35.