A thin rod of mass m and length l is suspended from one of its ends. It is set into oscillation about a horizontal axis. Its angular speed is `omega` while passing through its mean position. How high will its centre of mass rise from its lowest position ?

Options

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

Answer

To solve the problem of how high the center of mass of a thin rod rises from its lowest position when it is set into oscillation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**:
We have a thin rod of mass \( m \) and length \( l \) suspended from one of its ends. The rod is oscillating about a horizontal axis with an angular speed \( \omega \) at its mean position.

2. **Understand the Energy Conservation**:
We will use the principle of conservation of energy. The total mechanical energy (kinetic + potential) at the mean position will be equal to the total mechanical energy at the highest point of oscillation.

3. **Calculate Initial Kinetic Energy**:
At the mean position, the potential energy (PE) is zero (since we take this as the reference point), and the kinetic energy (KE) of the rod is given by:
\[ KE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the rod about the end. For a thin rod rotating about one end, the moment of inertia is:
\[ I = \frac{1}{3} m l^2 \]
Therefore, the initial kinetic energy can be expressed as:
\[ KE = \frac{1}{2} \left(\frac{1}{3} m l^2\right) \omega^2 = \frac{1}{6} m l^2 \omega^2 \]

4. **Calculate Potential Energy at Maximum Height**:
At the maximum height, the rod momentarily comes to rest, so all the kinetic energy is converted into potential energy. The potential energy at height \( h \) is given by:
\[ PE = mgh \]

5. **Set Up the Energy Conservation Equation**:
According to the conservation of energy:
\[ KE_{\text{initial}} = PE_{\text{final}} \]
Substituting the expressions we derived:
\[ \frac{1}{6} m l^2 \omega^2 = mgh \]

6. **Solve for Height \( h \)**:
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{6} l^2 \omega^2 = gh \]
Rearranging for \( h \):
\[ h = \frac{l^2 \omega^2}{6g} \]

### Final Result:
The height \( h \) that the center of mass rises from its lowest position is:
\[ h = \frac{l^2 \omega^2}{6g} \]