Question

Moment of inertia about an axis \( AB \) for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass 10 kg and radius \( R \) about an axis parallel to \( AB \) axis with separation of 3 m as shown in the figure below. The value of \( R \) is given as \( \sqrt{\frac{\alpha}{2}} \). The value of \( \alpha \) is _______.

Answer 1

Correct Answer: 15

Step 1: Moment of inertia of the rod.
The moment of inertia of the rod about an axis passing through one end and perpendicular to its length is given by: \[ I_{\text{rod}} = \frac{1}{3} M L^2, \] where \( M \) is the mass of the rod and \( L \) is the length of the rod. Substituting the given values \( M = 40 \, \text{kg} \) and \( L = 3 \, \text{m} \): \[ I_{\text{rod}} = \frac{1}{3} \times 40 \times (3)^2 = \frac{1}{3} \times 40 \times 9 = 120 \, \text{kg} \cdot \text{m}^2. \]
Step 2: Moment of inertia of the solid sphere.
The moment of inertia of a solid sphere about an axis passing through its center is: \[ I_{\text{sphere}} = \frac{2}{5} m R^2, \] where \( m \) is the mass of the sphere and \( R \) is the radius of the sphere. The moment of inertia about the parallel axis is given by the parallel axis theorem: \[ I_{\text{sphere, parallel}} = I_{\text{sphere}} + m d^2, \] where \( d \) is the distance between the center of the sphere and the new axis. In this case, \( d = 3 \, \text{m} \). Thus, \[ I_{\text{sphere, parallel}} = \frac{2}{5} m R^2 + m d^2. \] Substitute \( m = 10 \, \text{kg} \) and \( d = 3 \, \text{m} \): \[ I_{\text{sphere, parallel}} = \frac{2}{5} \times 10 \times R^2 + 10 \times (3)^2 = \frac{2}{5} \times 10 \times R^2 + 90. \] Simplifying: \[ I_{\text{sphere, parallel}} = 4 R^2 + 90. \]
Step 3: Setting the moments of inertia equal.
The problem states that the moments of inertia are equal, so we equate \( I_{\text{rod}} \) and \( I_{\text{sphere, parallel}} \): \[ 120 = 4 R^2 + 90. \] Solving for \( R^2 \): \[ 120 - 90 = 4 R^2, \] \[ 30 = 4 R^2, \] \[ R^2 = \frac{30}{4} = 7.5. \] Thus, \( R = \sqrt{7.5} \).
Step 4: Relating \( R \) to \( \alpha \).
We are given that \( R = \sqrt{\frac{\alpha}{2}} \), so: \[ \sqrt{\frac{\alpha}{2}} = \sqrt{7.5}. \] Squaring both sides: \[ \frac{\alpha}{2} = 7.5, \] \[ \alpha = 15. \] Thus, the value of \( \alpha \) is: \[ \boxed{15}. \]