A rod of length `L` is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod, When it is in verticle position is

Options

A. `sqrt((2g)/L`
B. `sqrt((3g)/L`
C. `sqrt(g/(2L))`
D. `sqrt(g/L)`

Explanation

To find the angular velocity of a rod of length \( L \) when it is in a vertical position after being released from a horizontal position, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the System
The rod is hinged at one end and initially held in a horizontal position. When released, it will fall under the influence of gravity and rotate about the hinge.

### Step 2: Identify the Initial and Final States
- **Initial State**: The rod is horizontal. The center of mass of the rod is at a height of \( \frac{L}{2} \) above the lowest point (the vertical position).
- **Final State**: The rod is in a vertical position. At this point, all the potential energy has been converted into rotational kinetic energy.

### Step 3: Calculate Initial Potential Energy
The potential energy \( U \) of the rod when it is horizontal can be calculated as:
\[ U_i = mgh \]
where \( h \) is the height of the center of mass from the lowest point. Since the center of mass is at \( \frac{L}{2} \) when the rod is horizontal:
\[ U_i = mg \left( \frac{L}{2} \right) \]

### Step 4: Calculate Final Kinetic Energy
When the rod is in the vertical position, all the potential energy has been converted into rotational kinetic energy \( K \):
\[ K_f = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the rod about the hinge. For a rod of length \( L \) hinged at one end, the moment of inertia is:
\[ I = \frac{1}{3} m L^2 \]

### Step 5: Apply Conservation of Energy
According to the conservation of energy:
\[ U_i = K_f \]
Substituting the expressions for potential energy and kinetic energy:
\[ mg \left( \frac{L}{2} \right) = \frac{1}{2} \left( \frac{1}{3} m L^2 \right) \omega^2 \]

### Step 6: Simplify the Equation
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g \left( \frac{L}{2} \right) = \frac{1}{6} L^2 \omega^2 \]
Multiply both sides by 6:
\[ 3gL = L^2 \omega^2 \]

### Step 7: Solve for Angular Velocity \( \omega \)
Divide both sides by \( L^2 \):
\[ \omega^2 = \frac{3g}{L} \]
Taking the square root gives:
\[ \omega = \sqrt{\frac{3g}{L}} \]

### Final Answer
The angular velocity of the rod when it is in the vertical position is:
\[ \omega = \sqrt{\frac{3g}{L}} \]