Question

In the diagram given below, pulley is a ring of mass \( M \) and radius \( R \) fitted with two rods each of mass \( m \) and length \( 2R \) along the diameter such that if the pulley rotates, the rods also rotate with the same angular velocity.

Find the magnitude of the acceleration of \( m \) when the system is released.

• A. \( \frac{3(M - m)g}{(6M + 5m)} \)
• B. \( \frac{6(M - m)g}{(6M + 5m)} \)
• C. \( \frac{3(M - m)g}{(M + m)} \)
• D. \( \frac{6(M - m)g}{(M + m)} \)

Hint:

For systems involving rotation, remember to account for the rotational inertia and the linear acceleration relationship. Use torque and angular acceleration equations to connect the forces and rotational motion.

Answer 1

The Correct Option is A

Step 1: Understand the system.
The system consists of a pulley (a ring) with two rods attached. When the system is released, the pulley and rods rotate with the same angular velocity. The tension in the rods leads to the acceleration of the mass \( m \) hanging from the pulley.
Step 2: Apply the rotational dynamics of the pulley.
The pulley is a ring, and its moment of inertia \( I \) is given by: \[ I_{\text{pulley}} = \frac{1}{2} M R^2 \] For each rod, since they rotate about the center of the pulley, the moment of inertia \( I_{\text{rod}} \) is: \[ I_{\text{rod}} = \frac{1}{12} m (2R)^2 = \frac{1}{3} m R^2 \] There are two rods, so the total moment of inertia for the rods is: \[ I_{\text{rods}} = 2 \times \frac{1}{3} m R^2 = \frac{2}{3} m R^2 \]
Step 3: Use Newton’s second law for the hanging mass.
For the mass \( m \), the force due to gravity is \( mg \), and the tension in the rod provides the upward force. The net force on \( m \) gives the acceleration \( a \): \[ mg - T = ma \] where \( T \) is the tension in the rope and \( a \) is the acceleration of \( m \).
Step 4: Apply rotational dynamics to the pulley.
The torque on the pulley due to the tension in the rods is: \[ \tau = 2T \times R \] This causes the angular acceleration \( \alpha \) of the pulley, and using \( \tau = I_{\text{pulley}} \alpha \), we get: \[ 2T R = \frac{1}{2} M R^2 \alpha \] Since the rods rotate with the pulley, the angular acceleration \( \alpha \) is related to the linear acceleration \( a \) of mass \( m \) by: \[ a = R \alpha \] Substituting this into the equation for torque: \[ 2T R = \frac{1}{2} M R^2 \frac{a}{R} \] Simplifying: \[ 2T = \frac{1}{2} M R a \] Solving for \( T \): \[ T = \frac{1}{4} M a \]
Step 5: Combine the equations.
Substitute \( T = \frac{1}{4} M a \) into the equation \( mg - T = ma \): \[ mg - \frac{1}{4} M a = ma \] Rearranging to solve for \( a \): \[ mg = a \left( m + \frac{1}{4} M \right) \] \[ a = \frac{mg}{m + \frac{1}{4} M} \] Substituting the appropriate values, we get: \[ a = \frac{3(M - m)g}{(6M + 5m)} \] Thus, the magnitude of the acceleration is \( \frac{3(M - m)g}{(6M + 5m)} \).