A uniform rod of length 2l is placed with one end in contact with the horizontal table and is then inclined at an angle `alpha` to the horizontal and allowed to fall. When it becomes horizontal, its angular velocity will be

Options

A. `omega = sqrt(((3g sin alpha)/(2l)))`
B. `omega = sqrt(((2l)/(3g sin alpha)))`
C. `omega = sqrt(((g sin alpha)/(l)))`
D. `omega = sqrt(((l)/(g sin alpha)))`

Answer

To find the angular velocity of a uniform rod of length \(2L\) when it falls from an inclined position at an angle \(\alpha\) to a horizontal position, we can use the work-energy theorem. Here’s a step-by-step solution:

### Step 1: Understand the System
A uniform rod of length \(2L\) is inclined at an angle \(\alpha\) to the horizontal. One end of the rod is in contact with a horizontal table. We need to determine the angular velocity (\(\omega\)) of the rod when it becomes horizontal.

### Step 2: Identify Forces and Work Done
When the rod is released, the only force doing work on the rod is its weight \(mg\) acting at the center of mass, which is located at a distance \(L\) from the pivot point (the end of the rod in contact with the table).

### Step 3: Calculate the Change in Height
Initially, the center of mass of the rod is at a height given by:
\[ h = L \sin \alpha \]
When the rod becomes horizontal, the center of mass is at height \(0\). The change in height (\(\Delta h\)) is:
\[ \Delta h = L \sin \alpha - 0 = L \sin \alpha \]

### Step 4: Work Done by Gravity
The work done by gravity as the rod falls is:
\[ W = mg \Delta h = mg (L \sin \alpha) \]

### Step 5: Relate Work Done to Kinetic Energy
According to the work-energy theorem, the work done on the rod is equal to its change in kinetic energy. Initially, the kinetic energy is zero (the rod starts from rest). The final kinetic energy when the rod is horizontal is given by:
\[ KE = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia of the rod about the pivot point. For a rod of length \(2L\) rotating about one end, the moment of inertia is:
\[ I = \frac{1}{3} m (2L)^2 = \frac{4}{3} m L^2 \]

### Step 6: Set Up the Equation
Setting the work done equal to the change in kinetic energy:
\[ mg (L \sin \alpha) = \frac{1}{2} \left(\frac{4}{3} m L^2\right) \omega^2 \]

### Step 7: Simplify the Equation
Cancel \(m\) from both sides:
\[ g (L \sin \alpha) = \frac{2}{3} L^2 \omega^2 \]
Now, multiply both sides by \(\frac{3}{2L^2}\):
\[ \frac{3g \sin \alpha}{2L} = \omega^2 \]

### Step 8: Solve for Angular Velocity
Taking the square root of both sides gives:
\[ \omega = \sqrt{\frac{3g \sin \alpha}{2L}} \]

### Final Answer
The angular velocity of the rod when it becomes horizontal is:
\[ \omega = \sqrt{\frac{3g \sin \alpha}{2L}} \]