A cord is wound around the circumference of wheel of radius r. The axis of the wheel is horizontal and `MI` is `I`. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be

Options

A. `sqrt((2gh)/(I+mr^(2)))`
B. `((2mgh)/(I+mr^(2)))^(1//2)`
C. `((2mgh)/(I+2mr^(2)))^(1//2)`
D. `sqrt(2gh)`

Explanation

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

### Step 1: Identify the initial and final states
Initially, the weight (mass \( m \)) is at a height \( h \) and is at rest. The potential energy at this point is given by:
\[ PE_{\text{initial}} = mgh \]
The wheel is also at rest, so its initial kinetic energy is zero.

### Step 2: Determine the final state after the weight has fallen
After the weight has fallen through a distance \( h \), it has a velocity \( v \) and the wheel has an angular velocity \( \omega \). The potential energy of the weight has been converted into kinetic energy.

### Step 3: Write the expression for final energy
The total energy at the final state consists of:
1. The kinetic energy of the falling mass:
\[ KE_{\text{mass}} = \frac{1}{2} mv^2 \]
2. The kinetic energy of the rotating wheel:
\[ KE_{\text{wheel}} = \frac{1}{2} I \omega^2 \]
Thus, the total final energy is:
\[ E_{\text{final}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

### Step 4: Apply the conservation of energy principle
According to the conservation of energy:
\[ PE_{\text{initial}} = E_{\text{final}} \]
This gives us:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

### Step 5: Relate linear and angular velocities
Since the cord does not slip on the wheel, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = r\omega \]
Substituting \( v \) in terms of \( \omega \) into the energy equation:
\[ mgh = \frac{1}{2} m (r\omega)^2 + \frac{1}{2} I \omega^2 \]
This simplifies to:
\[ mgh = \frac{1}{2} m r^2 \omega^2 + \frac{1}{2} I \omega^2 \]

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

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

### Step 8: Take the square root to find \( \omega \)
Taking the square root of both sides, we find:
\[ \omega = \sqrt{\frac{2mgh}{I + mr^2}} \]

### Conclusion
Thus, the angular velocity \( \omega \) of the wheel after the weight has fallen through a distance \( h \) is:
\[ \omega = \sqrt{\frac{2mgh}{I + mr^2}} \]