A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be :-

Options

A. `(2mgh)/(I+2mr^(2))`
B. `(2mgh)/(I+mr^(2))`
C. 2gh
D. `(2gh)/(I+mr^(2))`

Answer

To solve the problem, we will apply the principle of conservation of energy. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the system
We have a wheel of radius \( r \) and moment of inertia \( I \). A mass \( m \) is attached to a cord that is wound around the wheel. When the mass falls a distance \( h \), it causes the wheel to rotate.

### Step 2: Initial conditions
Initially, both the mass and the wheel are at rest. Therefore, the initial kinetic energy of the system is zero:
\[ KE_{\text{initial}} = 0 \]
The potential energy of the mass at height \( h \) is given by:
\[ PE_{\text{initial}} = mgh \]

### Step 3: Final conditions
After the mass has fallen a distance \( h \), it has kinetic energy due to its linear motion, and the wheel has kinetic energy due to its rotation. The final kinetic energy of the mass is:
\[ KE_{\text{block}} = \frac{1}{2} mv^2 \]
The final kinetic energy of the wheel is:
\[ KE_{\text{wheel}} = \frac{1}{2} I \omega^2 \]
The potential energy of the mass at the new height (which is now \( 0 \) since it has fallen \( h \)) is:
\[ PE_{\text{final}} = 0 \]

### Step 4: Apply conservation of energy
According to the conservation of energy, the initial total energy equals the final total energy:
\[ PE_{\text{initial}} + KE_{\text{initial}} = PE_{\text{final}} + KE_{\text{block}} + KE_{\text{wheel}} \]
Substituting the values we have:
\[ mgh + 0 = 0 + \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
This simplifies to:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

### Step 5: Relate linear and angular velocities
The linear velocity \( v \) of the falling mass is related to the angular velocity \( \omega \) of the wheel by the equation:
\[ v = r \omega \]
Substituting this into our energy equation gives:
\[ mgh = \frac{1}{2} m(r \omega)^2 + \frac{1}{2} I \omega^2 \]
This simplifies to:
\[ mgh = \frac{1}{2} m r^2 \omega^2 + \frac{1}{2} I \omega^2 \]

### Step 6: Factor out \( \omega^2 \)
Factoring out \( \frac{1}{2} \omega^2 \) from the right side:
\[ mgh = \frac{1}{2} \omega^2 (mr^2 + I) \]

### Step 7: Solve for \( \omega^2 \)
Rearranging the equation to solve for \( \omega^2 \):
\[ \omega^2 = \frac{2mgh}{mr^2 + I} \]

### Final Answer
The square of the angular velocity of the wheel after the mass has fallen through a distance \( h \) is:
\[ \omega^2 = \frac{2mgh}{I + mr^2} \]