A rod of mass m and length L , lying horizontally is free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts.

Explanation

To solve the problem, we need to determine the angle rotated by the rod during a time \( t \) after a constant horizontal force \( F \) is applied. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the Torque
The torque \( \tau \) acting on the rod due to the force \( F \) can be calculated using the formula:
\[ \tau = F \cdot d \]
where \( d \) is the distance from the axis of rotation to the point where the force is applied. In this case, \( d = \frac{L}{4} \). Therefore, the torque is:
\[ \tau = F \cdot \frac{L}{4} \]

### Step 2: Calculate the Moment of Inertia
The moment of inertia \( I \) of a uniform rod about its center is given by:
\[ I = \frac{1}{12} m L^2 \]

### Step 3: Relate Torque to Angular Acceleration
Using the relation between torque and angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
Substituting the values we have:
\[ F \cdot \frac{L}{4} = \frac{1}{12} m L^2 \cdot \alpha \]

### Step 4: Solve for Angular Acceleration
Rearranging the equation to solve for \( \alpha \):
\[ \alpha = \frac{F \cdot \frac{L}{4}}{\frac{1}{12} m L^2} \]
This simplifies to:
\[ \alpha = \frac{F \cdot 3}{m L} \]

### Step 5: Use Angular Kinematics to Find the Angle Rotated
Since the initial angular velocity \( \omega_0 = 0 \) (the rod starts from rest), we can use the angular displacement formula:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
Substituting \( \omega_0 = 0 \):
\[ \theta = \frac{1}{2} \alpha t^2 \]
Now substituting the value of \( \alpha \):
\[ \theta = \frac{1}{2} \left(\frac{3F}{mL}\right) t^2 \]
This simplifies to:
\[ \theta = \frac{3Ft^2}{2mL} \]

### Final Result
Thus, the angle rotated by the rod during the time \( t \) is:
\[ \theta = \frac{3Ft^2}{2mL} \]