A wheel whose radius is R and moment of inertia about its own axis is I, rotates freely about its own axis. A rope is wrapped on the wheel. A body of mass m is suspended from the free end of the rope. The body is released from rest. The velocity of the body after falling a distance h would be:

Options

A. `((mgh)/(I))^(1//2)`
B. `((2mgh)/(m+I))^(1//2)`
C. `((2mgh)/(m+I//r^(2)))^(1//2)`
D. `((m+I)/(2mgh))^(1//2)`

Answer

To solve the problem, we need to analyze the motion of the body and the wheel using Newton's laws and the rotational dynamics principles. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Mass
When the mass \( m \) is released, two forces act on it:
- The gravitational force \( mg \) acting downward.
- The tension \( T \) in the rope acting upward.

Using Newton's second law for the mass \( m \), we can write the equation of motion:
\[ mg - T = ma \quad (1) \]
where \( a \) is the linear acceleration of the mass.

### Step 2: Relate Linear Acceleration to Angular Acceleration
The wheel rotates due to the tension in the rope. The torque \( \tau \) acting on the wheel can be expressed as:
\[ \tau = T \cdot R \]
where \( R \) is the radius of the wheel. The torque is also related to the angular acceleration \( \alpha \) by:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia of the wheel.

Since the linear acceleration \( a \) of the mass is related to the angular acceleration \( \alpha \) of the wheel by:
\[ \alpha = \frac{a}{R} \quad (2) \]
we can substitute this into the torque equation:
\[ T \cdot R = I \cdot \frac{a}{R} \]
Rearranging gives:
\[ T = \frac{I a}{R^2} \quad (3) \]

### Step 3: Substitute Equation (3) into Equation (1)
Now we can substitute \( T \) from equation (3) into equation (1):
\[ mg - \frac{I a}{R^2} = ma \]
Rearranging this gives:
\[ mg = ma + \frac{I a}{R^2} \]
Factoring out \( a \) from the right side:
\[ mg = a \left( m + \frac{I}{R^2} \right) \]
Now, solving for \( a \):
\[ a = \frac{mg}{m + \frac{I}{R^2}} \quad (4) \]

### Step 4: Use Kinematic Equation to Find Final Velocity
We need to find the velocity \( v \) of the mass after falling a distance \( h \). We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Since the initial velocity \( u = 0 \), this simplifies to:
\[ v^2 = 2as \]
Substituting \( s = h \) and using equation (4) for \( a \):
\[ v^2 = 2 \left( \frac{mg}{m + \frac{I}{R^2}} \right) h \]
Thus, we have:
\[ v = \sqrt{2h \cdot \frac{mg}{m + \frac{I}{R^2}}} \]

### Step 5: Final Expression for Velocity
This gives us the final expression for the velocity of the body after falling a distance \( h \):
\[ v = \sqrt{\frac{2mgh}{m + \frac{I}{R^2}}} \]

### Summary
The velocity of the body after falling a distance \( h \) is given by:
\[ v = \sqrt{\frac{2mgh}{m + \frac{I}{R^2}}} \]