In the figure shown a uniform rod of length l and mass m is kept at rest in horizontal position on an elevated edge. The value of x is such that the rod will have maximum angular acceleration `alpha`, as soon as it is set free.

Options

A. x is equal to `(l)/(2 sqrt(3))`
B. `alpha` is equal to `(g sqrt(3))/(2l)`
C. `alpha` is equal to `(g sqrt(3))/(l)`
D. x is equal to `(l)/( sqrt(3))`

Answer

To solve the problem of finding the value of \( x \) such that the rod will have maximum angular acceleration \( \alpha \) when set free, we can follow these steps:

### Step 1: Understand the Setup
We have a uniform rod of length \( l \) and mass \( m \) resting horizontally on an elevated edge. The edge is at a distance \( x \) from one end of the rod. When the rod is released, it will pivot about the edge.

### Step 2: Identify Forces and Torque
When the rod is released, the force acting on it due to gravity is \( mg \) (where \( g \) is the acceleration due to gravity). The torque \( \tau \) about the pivot point (the edge) can be calculated as:
\[ \tau = r \times F \]
where \( r \) is the distance from the pivot to the center of mass of the rod, and \( F \) is the force acting downwards (which is \( mg \)).

### Step 3: Calculate the Distance to the Center of Mass
The center of mass of the rod is located at a distance of \( \frac{l}{2} \) from one end. Therefore, the distance from the pivot point to the center of mass is:
\[ r = \frac{l}{2} - x \]

### Step 4: Write the Expression for Torque
The torque about the pivot point is given by:
\[ \tau = mg \left(\frac{l}{2} - x\right) \]

### Step 5: Moment of Inertia
The moment of inertia \( I \) of the rod about the pivot point can be calculated using the parallel axis theorem:
\[ I = I_{cm} + md^2 \]
where \( I_{cm} = \frac{ml^2}{12} \) is the moment of inertia about the center of mass, and \( d = \left(\frac{l}{2} - x\right) \) is the distance from the center of mass to the pivot point. Thus,
\[ I = \frac{ml^2}{12} + m\left(\frac{l}{2} - x\right)^2 \]

### Step 6: Relate Torque and Angular Acceleration
According to Newton's second law for rotation:
\[ \tau = I \alpha \]
Substituting the expressions we have:
\[ mg \left(\frac{l}{2} - x\right) = \left(\frac{ml^2}{12} + m\left(\frac{l}{2} - x\right)^2\right) \alpha \]

### Step 7: Maximize Angular Acceleration
To find the maximum angular acceleration \( \alpha \), we need to differentiate the expression with respect to \( x \) and set the derivative to zero:
\[ \frac{d\alpha}{dx} = 0 \]
This will give us the value of \( x \) that maximizes \( \alpha \).

### Step 8: Solve for \( x \)
After performing the differentiation and setting it to zero, we find:
\[ x = \frac{l}{2\sqrt{3}} \]

### Conclusion
The value of \( x \) that gives the maximum angular acceleration \( \alpha \) is:
\[ x = \frac{l}{2\sqrt{3}} \]