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 centre of mass rises to a maximum height of -

Options

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

Answer

To solve the problem, we will use the principle of conservation of energy. The rotational kinetic energy of the rod when it is swinging will be converted into gravitational potential energy when the center of mass rises to its maximum height.

### Step-by-step Solution:

1. **Identify the Energy Forms**:
- The initial energy of the system when the rod is swinging is its rotational kinetic energy.
- The final energy when the center of mass reaches its maximum height is its gravitational potential energy.

2. **Write the Expression for Rotational Kinetic Energy**:
- The rotational kinetic energy (KE) of the rod can be expressed as:
\[ KE = \frac{1}{2} I \omega^2 \]
- Here, \(I\) is the moment of inertia of the rod about the end, and \(\omega\) is the angular speed.

3. **Calculate the Moment of Inertia**:
- The moment of inertia \(I\) for a thin uniform rod of length \(l\) and mass \(m\) about an axis through one end is given by:
\[ I = \frac{1}{3} m l^2 \]

4. **Substitute the Moment of Inertia into the Kinetic Energy Equation**:
- Substitute \(I\) into the kinetic energy equation:
\[ KE = \frac{1}{2} \left(\frac{1}{3} m l^2\right) \omega^2 = \frac{1}{6} m l^2 \omega^2 \]

5. **Write the Expression for Gravitational Potential Energy**:
- The gravitational potential energy (PE) when the center of mass rises to a height \(h\) is given by:
\[ PE = mgh \]

6. **Apply Conservation of Energy**:
- According to the conservation of energy, the initial kinetic energy will equal the potential energy at the maximum height:
\[ \frac{1}{6} m l^2 \omega^2 = mgh \]

7. **Cancel the Mass \(m\)**:
- Since \(m\) appears on both sides of the equation, we can cancel it out:
\[ \frac{1}{6} l^2 \omega^2 = gh \]

8. **Solve for Height \(h\)**:
- Rearranging the equation to solve for \(h\):
\[ h = \frac{l^2 \omega^2}{6g} \]

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