A uniform rod of length L and mass M is held vertical, with its bottom end pivoted to the floor. The rod falls under gravity, freely turning about the pivot. If acceleration due to gravity is g, what is the instantaneous angular speed of the rod when it makes an angle `60^(@)` with the vertical

Options

A. `((g)/(L))^(1//2)`
B. `((3g)/(4L))^(1//2)`
C. `((3sqrt(3)g)/(2L))^(1//2)`
D. `((3g)/(2L))^(1//2)`

Explanation

To solve the problem of finding the instantaneous angular speed of a uniform rod when it makes an angle of 60 degrees with the vertical, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Initial and Final States
Initially, the rod is vertical, and its center of mass is at a height of \( \frac{L}{2} \) from the pivot. When the rod falls and makes an angle of 60 degrees with the vertical, its center of mass is at a lower height.

### Step 2: Calculate the Initial Potential Energy
The initial potential energy (PE_initial) of the rod when it is vertical can be calculated as:
\[ PE_{\text{initial}} = M g \left(\frac{L}{2}\right) \]

### Step 3: Calculate the Final Height of the Center of Mass
When the rod makes an angle of 60 degrees with the vertical, the height of the center of mass from the pivot is:
\[ h_{\text{final}} = \frac{L}{2} \cos(60^\circ) = \frac{L}{2} \cdot \frac{1}{2} = \frac{L}{4} \]

### Step 4: Calculate the Final Potential Energy
The final potential energy (PE_final) when the rod is at 60 degrees is:
\[ PE_{\text{final}} = M g \left(\frac{L}{4}\right) \]

### Step 5: Set Up the Conservation of Energy Equation
According to the conservation of energy:
\[ PE_{\text{initial}} + KE_{\text{initial}} = PE_{\text{final}} + KE_{\text{final}} \]
Since the rod starts from rest, \( KE_{\text{initial}} = 0 \). Thus, we have:
\[ M g \left(\frac{L}{2}\right) = M g \left(\frac{L}{4}\right) + \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the rod about the pivot, which is given by:
\[ I = \frac{1}{3} M L^2 \]

### Step 6: Substitute the Moment of Inertia
Substituting \( I \) into the equation:
\[ M g \left(\frac{L}{2}\right) = M g \left(\frac{L}{4}\right) + \frac{1}{2} \left(\frac{1}{3} M L^2\right) \omega^2 \]

### Step 7: Simplify the Equation
Cancel \( M \) from all terms (assuming \( M \neq 0 \)):
\[ g \left(\frac{L}{2}\right) = g \left(\frac{L}{4}\right) + \frac{1}{6} L^2 \omega^2 \]
Now, simplifying further:
\[ g \left(\frac{L}{2} - \frac{L}{4}\right) = \frac{1}{6} L^2 \omega^2 \]
\[ g \left(\frac{L}{4}\right) = \frac{1}{6} L^2 \omega^2 \]

### Step 8: Solve for Angular Speed \( \omega \)
Rearranging gives:
\[ \omega^2 = \frac{6g}{L} \cdot \frac{L}{4} \]
\[ \omega^2 = \frac{6g}{4} = \frac{3g}{2} \]
Taking the square root:
\[ \omega = \sqrt{\frac{3g}{2}} \]

### Final Answer
The instantaneous angular speed of the rod when it makes an angle of 60 degrees with the vertical is:
\[ \omega = \sqrt{\frac{3g}{2}} \]