Question

A cockroach of mass \( m \) is moving with a velocity \( v \) in the anticlockwise direction on the rim of a disc of radius \( R \). The moment of inertia of the disc about the axis is \( I \) and it is rotating in a clockwise direction with an angular velocity \( \omega \). If the cockroach stops moving, the angular velocity of the disc will be:

• A. \(\frac{I\omega + mR^2}{I + mR^2}\)
• B. \(\frac{I\omega + mRv}{I + mR^2}\)
• C. \(\frac{I\omega - mRv}{I + mR^2}\)
• D. \(\frac{I\omega - mRv}{I}\)

Hint:

When calculating systems with moving parts on rotating bodies, always consider both linear and angular momentum contributions.

Answer 1

The Correct Option is C

Step 1: Apply conservation of angular momentum.
Before the cockroach stops, the system's angular momentum is: \[ L_{{initial}} = I\omega - mRv \] The negative sign arises because the cockroach’s velocity direction (anticlockwise) opposes the disc's rotation direction (clockwise). 
Step 2: After the cockroach stops.
The cockroach's angular momentum becomes zero, and the disc's angular momentum must now equal the initial total angular momentum: \[ L_{{final}} = I\omega_{{final}} \] Step 3: Solve for the final angular velocity \(\omega_{{final}}\).
Using the conservation of angular momentum: \[ I\omega - mRv = I\omega_{{final}} + mR^2\omega_{{final}} \] \[ \omega_{{final}} = \frac{I\omega - mRv}{I + mR^2} \]

Answer 2

To solve the problem, we need to determine the angular velocity of the disc when the cockroach stops moving.

1. Understanding the Scenario:
We have a cockroach of mass \( m \) moving with velocity \( u \) on the rim of a disc of radius \( R \). The disc has a moment of inertia \( I \) about the axis and is rotating with angular velocity \( \omega \). When the cockroach stops moving, we need to find the new angular velocity of the disc.

2. Applying Conservation of Angular Momentum:
Since there are no external torques acting on the system, the total angular momentum of the system before and after the cockroach stops moving is conserved.

The initial angular momentum of the system is the sum of the angular momentum of the cockroach and the disc: \[ L_{\text{initial}} = m \cdot u \cdot R + I \cdot \omega \] where \( m \cdot u \cdot R \) is the angular momentum of the cockroach, and \( I \cdot \omega \) is the angular momentum of the disc. After the cockroach stops moving, only the disc is rotating, so the final angular momentum is: \[ L_{\text{final}} = I \cdot \omega_{\text{final}} \] where \( \omega_{\text{final}} \) is the angular velocity of the disc after the cockroach stops. Since angular momentum is conserved: \[ m \cdot u \cdot R + I \cdot \omega = I \cdot \omega_{\text{final}} \] Solving for \( \omega_{\text{final}} \): \[ \omega_{\text{final}} = \frac{I \cdot \omega + m \cdot u \cdot R}{I + m \cdot R^2} \]

3. Final Answer:
The correct formula for the final angular velocity of the disc is:

Final Answer:
The correct option is (C) \( \frac{I \cdot \omega + m \cdot u \cdot R}{I + m \cdot R^2} \).