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 when oscillating about a horizontal axis, we can use the principle of conservation of mechanical energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
The rod is suspended from one of its ends and oscillates about that point. The center of mass of the rod is located at a distance of \( \frac{l}{2} \) from the pivot point.

### Step 2: Identify Initial Conditions
At the mean position, the rod has an angular speed \( \omega \). The initial kinetic energy (KE) of the rod can be calculated using the moment of inertia \( I \) about the pivot point.

### Step 3: Calculate the Moment of Inertia
The moment of inertia \( I \) of a thin rod about an end is given by:
\[ I = \frac{1}{3} m l^2 \]

### Step 4: Calculate Initial Kinetic Energy
The initial kinetic energy when the rod is at the mean position is:
\[ KE_i = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{3} m l^2\right) \omega^2 = \frac{1}{6} m l^2 \omega^2 \]

### Step 5: Set Up Energy Conservation Equation
According to the conservation of mechanical energy:
\[ \text{Initial Potential Energy} + \text{Initial Kinetic Energy} = \text{Final Potential Energy} + \text{Final Kinetic Energy} \]
At the mean position, the potential energy (PE) is zero, and at the extreme position, the kinetic energy is zero. Thus, we have:
\[ 0 + \frac{1}{6} m l^2 \omega^2 = mgh + 0 \]

### Step 6: Solve for Height \( h \)
Rearranging the equation gives:
\[ \frac{1}{6} m l^2 \omega^2 = mgh \]
Dividing both sides by \( m \) and rearranging for \( h \):
\[ h = \frac{l^2 \omega^2}{6g} \]

### Step 7: Conclusion
The height \( h \) that the center of mass rises from its lowest position is:
\[ h = \frac{l^2 \omega^2}{6g} \]