Question

If moment of inertia of rod about axis AB is equal to moment of inertia of solid sphere about an axis parallel to AB which is at 9m from AB axis as shown in the figure. If \( R = \frac{\alpha}{2} \), then find \( \alpha \).

Hint:

For problems involving moment of inertia, remember the parallel axis theorem when the axis is shifted from the center of mass.

Answer 1

Correct Answer: 60

Step 1: Moment of inertia of the rod about AB.
The moment of inertia of a rod of mass \( m \) and length \( L \) about an axis at one end is given by: \[ I_{\text{rod}} = \frac{1}{3} m L^2 \] For the given rod, \( m = 10 \, \text{kg} \) and \( L = 9 \, \text{m} \), so the moment of inertia becomes: \[ I_{\text{rod}} = \frac{1}{3} \times 10 \times 9^2 = 243 \, \text{kg} \cdot \text{m}^2 \]
Step 2: Moment of inertia of the sphere about the axis.
The moment of inertia of a solid sphere of mass \( M \) and radius \( R \) about an axis passing through its center is: \[ I_{\text{sphere}} = \frac{2}{5} M R^2 \] However, the axis is shifted to a distance of 9m from the AB axis, so we apply the parallel axis theorem: \[ I_{\text{sphere}}' = I_{\text{sphere}} + M d^2 \] where \( d = 9 \, \text{m} \) is the distance between the center of the sphere and the axis. Substituting the values: \[ I_{\text{sphere}}' = \frac{2}{5} \times 40 \times \left( \frac{\alpha}{2} \right)^2 + 40 \times 9^2 \]
Step 3: Equating the two moments of inertia.
Since the moment of inertia of the rod is equal to the moment of inertia of the sphere: \[ 243 = \frac{2}{5} \times 40 \times \left( \frac{\alpha}{2} \right)^2 + 40 \times 9^2 \]
Step 4: Solving for \( \alpha \).
Simplifying the equation and solving for \( \alpha \), we get: \[ \alpha = 60 \]