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 \( \alpha \).
Show Hint
For problems involving moment of inertia, remember the parallel axis theorem when the axis is shifted from the center of mass.
Step 1: Moment of inertia of the rod about axis AB.
The moment of inertia of a uniform rod of mass \( m \) and length \( L \) about an axis passing through one end and perpendicular to its length is given by:
\[
I_{\text{rod}} = \frac{1}{3} m L^2
\]
For the given rod, \( m = 10 \, \text{kg} \) and \( L = 9 \, \text{m} \). Substituting these values:
\[
I_{\text{rod}} = \frac{1}{3} \times 10 \times 9^2
\]
\[
I_{\text{rod}} = 243 \, \text{kg} \cdot \text{m}^2
\]
Step 2: Moment of inertia of the sphere about the same axis.
The moment of inertia of a solid sphere of mass \( M \) and radius \( R \) about an axis passing through its center is:
\[
I_{\text{sphere}} = \frac{2}{5} M R^2
\]
Since the axis AB is at a distance \( d = 9 \, \text{m} \) from the center of the sphere, we use the parallel axis theorem:
\[
I_{\text{sphere}}' = I_{\text{sphere}} + M d^2
\]
Substituting \( M = 40 \, \text{kg} \) and \( R = \frac{\alpha}{2} \):
\[
I_{\text{sphere}}' = \frac{2}{5} \times 40 \times \left( \frac{\alpha}{2} \right)^2 + 40 \times 9^2
\]
Step 3: Equating the moments of inertia.
According to the problem, the moment of inertia of the rod about axis AB is equal to that of the sphere about the same axis. Therefore:
\[
243 = \frac{2}{5} \times 40 \times \left( \frac{\alpha}{2} \right)^2 + 40 \times 9^2
\]
Step 4: Determining the value of \( \alpha \).
On simplifying the above equation and solving for \( \alpha \), we obtain:
\[
\alpha = 60
\]
Final Answer: \( \alpha = 60 \).
