Equate Moment of Inertia of Rod and Solid Sphere

To find the value of $\alpha$, we determine the moment of inertia for both objects and set them equal.

The moment of inertia of the rod of mass $M = 40\text{ kg}$ and length $L = 3\text{ m}$ about the axis $AB$ passing through its end is: $I_{\text{rod}} = \frac{1}{3} M L^2 = \frac{1}{3} \times 40 \times 3^2 = 120\text{ kg}\cdot\text{m}^2$

Equating this to the moment of inertia of the solid sphere of mass $M_{\text{sphere}} = 10\text{ kg}$ and radius $R$: $I_{\text{sphere}} = \frac{2}{5} M_{\text{sphere}} R^2 = \frac{2}{5} \times 10 \times R^2 = 4R^2$

Setting $I_{\text{rod}} = I_{\text{sphere}}$: $120 = 4R^2 \implies R^2 = 30 \implies R = \sqrt{30} = \sqrt{\frac{60}{2}}$

Comparing this with $R = \sqrt{\frac{\alpha}{2}}$, we get: $\alpha = 60$