A thin bar of mass `M` and length `L` is free to rotate about a fixed horizontal axis through a point at its end. The bar is brought to a horizontal position and then released. The axis is perpendicular to the rod. The angular velocity when it reaches the lowest point is

Options

A. directly proportional to its length and inversely proportional to its mass
B. independent of mass and inversely proportional to the square root of its length
C. dependent only upon the acceleration due to gravity and the length of the bar
D. directly proportional to its length and inversely proportional to the acceleration due to gravity

Explanation

To solve the problem, we will use the principle of conservation of energy and the work-energy theorem. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial and Final Conditions
- The bar is initially horizontal and then released. At this point, it has potential energy and no kinetic energy.
- When the bar reaches the lowest point (vertical position), it has kinetic energy due to its rotation about the fixed axis.

### Step 2: Calculate the Work Done by Gravity
- The center of mass of the bar is located at a distance of \( \frac{L}{2} \) from the pivot point.
- The change in height of the center of mass when the bar moves from horizontal to vertical is \( \frac{L}{2} \).
- The work done by gravity \( W \) can be calculated as:
\[ W = mgh = mg \left( \frac{L}{2} \right) = \frac{mgL}{2} \]

### Step 3: Apply the Work-Energy Theorem
- According to the work-energy theorem, the work done by gravity is equal to the change in kinetic energy:
\[ W = \Delta KE \]
- Since the initial kinetic energy is zero (the bar is released from rest), we have:
\[ \frac{mgL}{2} = KE_{final} - KE_{initial} = KE_{final} - 0 = KE_{final} \]

### Step 4: Express the Final Kinetic Energy
- The final kinetic energy of the rotating bar can be expressed as:
\[ KE_{final} = \frac{1}{2} I \omega^2 \]
- The moment of inertia \( I \) of a thin bar about an end is given by:
\[ I = \frac{1}{3} mL^2 \]
- Therefore, substituting for \( KE_{final} \):
\[ \frac{mgL}{2} = \frac{1}{2} \left( \frac{1}{3} mL^2 \right) \omega^2 \]

### Step 5: Simplify the Equation
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{gL}{2} = \frac{1}{6} L^2 \omega^2 \]
- Rearranging gives:
\[ \omega^2 = \frac{3g}{L} \]

### Step 6: Solve for Angular Velocity
- Taking the square root of both sides, we find:
\[ \omega = \sqrt{\frac{3g}{L}} \]

### Final Answer
The angular velocity \( \omega \) when the bar reaches the lowest point is:
\[ \omega = \sqrt{\frac{3g}{L}} \]