A uniform rod of length 2L is placed with one end in contact with the horizontal and is then inclined at an angle `alpha` to the horizontal and allowed to fall without slipping at contact point. When it becomes horizontal, its angular velocity will be

Options

A. `omega=sqrt((3gsinalpha)/(2L))`
B. `omega=sqrt((2L)/(3gsinalpha))`
C. `omega=sqrt((6gsinalpha)/(L))`
D. `omega=sqrt((L)/(g sin alpha))`

Explanation

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

### Step 1: Determine the Initial Potential Energy
When the rod is inclined at an angle \(\alpha\), the center of mass of the rod is at a height \(h\) above the horizontal. The height \(h\) can be calculated as:
\[ h = L \sin \alpha \]
Thus, the initial potential energy (PE_initial) of the rod is given by:
\[ PE_{\text{initial}} = mgh = mg(L \sin \alpha) \]

### Step 2: Determine the Final Kinetic Energy
When the rod becomes horizontal, all the potential energy will have converted into kinetic energy. The kinetic energy (KE_final) of the rod when it is horizontal can be expressed as:
\[ KE_{\text{final}} = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia of the rod about the pivot point (one end) and \(\omega\) is the angular velocity.

### Step 3: Calculate the Moment of Inertia
For a uniform rod of length \(2L\) about one end, the moment of inertia \(I\) is given by:
\[ I = \frac{1}{3} m (2L)^2 = \frac{4}{3} m L^2 \]

### Step 4: Apply Conservation of Energy
According to the conservation of energy, the initial potential energy must equal the final kinetic energy:
\[ mg(L \sin \alpha) = \frac{1}{2} I \omega^2 \]
Substituting the expression for \(I\):
\[ mg(L \sin \alpha) = \frac{1}{2} \left(\frac{4}{3} m L^2\right) \omega^2 \]

### Step 5: Simplify the Equation
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ g(L \sin \alpha) = \frac{2}{3} L^2 \omega^2 \]
Now, we can simplify further:
\[ g \sin \alpha = \frac{2}{3} L \omega^2 \]

### Step 6: Solve for Angular Velocity \(\omega\)
Rearranging the equation to solve for \(\omega^2\):
\[ \omega^2 = \frac{3g \sin \alpha}{2L} \]
Taking the square root to find \(\omega\):
\[ \omega = \sqrt{\frac{3g \sin \alpha}{2L}} \]

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