A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is `omega`. Its cenre of mass rises to a maximum height of :

Options

A. `(1 l omega)/(6 g)`
B. `(1 l^2 omega^2)/(g)`
C. `(1 l^2 omega^2)/(6 g)`
D. `1(l^2 omega^2)/(3g)`

Explanation

To solve the problem of finding the maximum height to which the center of mass of a thin uniform rod rises when swinging about a horizontal axis, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the System
We have a thin uniform rod of length \( l \) and mass \( m \) swinging about a horizontal axis at one end. The rod is initially in a vertical position and swings down to a horizontal position.

### Step 2: Identify the Initial and Final States
- **Initial State**: The rod is vertical, and its center of mass is at a height of \( \frac{l}{2} \) from the axis of rotation.
- **Final State**: The rod swings down to a horizontal position, and we want to find the height \( h \) to which the center of mass rises after swinging.

### Step 3: Apply Conservation of Energy
The mechanical energy of the system is conserved. The initial potential energy when the rod is vertical will convert into kinetic energy at the lowest point and then into potential energy when the rod swings up to its maximum height.

1. **Initial Potential Energy (PE_initial)**:
\[ PE_{\text{initial}} = mgh_{\text{initial}} = mg\left(\frac{l}{2}\right) \]

2. **Kinetic Energy at the Lowest Point (KE)**:
The kinetic energy when the rod is horizontal can be expressed in terms of its angular speed \( \omega \):
\[ KE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the rod about the end:
\[ I = \frac{1}{3} ml^2 \]
Therefore,
\[ KE = \frac{1}{2} \left(\frac{1}{3} ml^2\right) \omega^2 = \frac{1}{6} ml^2 \omega^2 \]

### Step 4: Set Up the Energy Conservation Equation
At the highest point, all kinetic energy converts back into potential energy:
\[ KE = PE_{\text{final}} \]
The potential energy at the maximum height \( h \) is:
\[ PE_{\text{final}} = mg(h + \frac{l}{2}) \]
Setting the initial potential energy equal to the kinetic energy gives:
\[ mg\left(\frac{l}{2}\right) = \frac{1}{6} ml^2 \omega^2 \]

### Step 5: Solve for the Maximum Height \( h \)
Rearranging the equation:
\[ g\left(\frac{l}{2}\right) = \frac{1}{6} l^2 \omega^2 \]
\[ h + \frac{l}{2} = \frac{l^2 \omega^2}{6g} \]
\[ h = \frac{l^2 \omega^2}{6g} - \frac{l}{2} \]
To express \( h \) in terms of \( l \), we can simplify:
\[ h = \frac{l^2 \omega^2}{6g} - \frac{3l}{6} = \frac{l^2 \omega^2 - 3lg}{6g} \]

### Conclusion
The maximum height to which the center of mass rises is:
\[ h = \frac{l^2 \omega^2}{6g} - \frac{l}{2} \]