Question

A cord is wound round the circumference of a wheel of radius \( r \). The axis of the wheel is horizontal and the moment of inertia about it is \( I \). A block of mass \( m \) is attached to the free end of the cord, initially at rest. When the wheel rotates and the block moves vertically downwards through distance \( h \), the angular velocity of the wheel will be

• A. \( \left( \frac{2gh}{I+mr} \right)^{\frac{1}{2}} \)
• B. \( \left( \frac{2mgh}{I+mr^2} \right)^{\frac{1}{2}} \)
• C. \( \left( \frac{2mgh}{I+2m} \right)^{\frac{1}{2}} \)
• D. \( \left( \frac{2gh}{I+mr} \right)^{\frac{1}{2}} \)

Hint:

In rotational motion problems, apply the conservation of mechanical energy and use the relationship between linear and angular quantities to solve for the desired variable.

Answer 1

The Correct Option is B

Step 1: Understanding the energy conservation principle.
The total mechanical energy at the start is purely potential energy, \( mgh \). When the block moves downwards, the energy is split into rotational kinetic energy of the wheel and the translational kinetic energy of the block. The total energy is conserved.
Step 2: Setting up the equation.
The rotational kinetic energy of the wheel is \( \frac{1}{2}I\omega^2 \), and the translational kinetic energy of the block is \( \frac{1}{2}mv^2 \). The work done by gravity is \( mgh \). Using the relationship between linear and angular velocity \( v = r\omega \), we get: \[ mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mr^2\omega^2 \] Solving for \( \omega \), we get: \[ \omega = \left( \frac{2mgh}{I + mr^2} \right)^{\frac{1}{2}} \] Step 3: Conclusion.
Thus, the angular velocity of the wheel is \( \left( \frac{2mgh}{I + mr^2} \right)^{\frac{1}{2}} \), corresponding to option (B).