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: \[ I\omega - mRv = I\omega_{{final}} + mR^2\omega_{{final}} \] \[ \omega_{{final}} = \frac{I\omega - mRv}{I + mR^2} \]