A rod of length `L`, hinged at the bottom is held vertically and then allowed to fall, the linear velocity of its top when it hits the floor is

Options

A. `sqrt(2gL)`
B. `sqrt((2g)/(L))`
C. `sqrt(3gL)`
D. `sqrt((3g)/(L))`

Explanation

To find the linear velocity of the top of the rod when it hits the floor, we can use the principles of rotational motion and energy conservation. Here’s a step-by-step solution:

### Step 1: Understand the System
The rod is hinged at the bottom and falls under the influence of gravity. When it falls, it rotates about the hinge point at the bottom.

### Step 2: Identify the Initial and Final States
- **Initial State**: The rod is vertical and at rest. The potential energy is maximum, and kinetic energy is zero.
- **Final State**: The rod is horizontal just before it hits the floor. The potential energy is zero, and kinetic energy is at its maximum.

### Step 3: Calculate the Initial Potential Energy
The center of mass of the rod is located at a distance of \( \frac{L}{2} \) from the hinge. The initial height of the center of mass is \( \frac{L}{2} \), so the initial potential energy (PE_initial) is given by:
\[ PE_{\text{initial}} = mgh = mg\left(\frac{L}{2}\right) \]
where \( m \) is the mass of the rod and \( g \) is the acceleration due to gravity.

### Step 4: Calculate the Final Kinetic Energy
When the rod is horizontal, all the potential energy has converted into kinetic energy. The kinetic energy (KE) of the rod 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 hinge and \( \omega \) is the angular velocity.

For a rod rotating about one end, the moment of inertia \( I \) is:
\[ I = \frac{1}{3} mL^2 \]

### Step 5: Relate Angular Velocity to Linear Velocity
The linear velocity \( v \) of the top of the rod is related to the angular velocity \( \omega \) by:
\[ v = \omega L \]

### Step 6: Apply Conservation of Energy
Setting the initial potential energy equal to the final kinetic energy:
\[ mg\left(\frac{L}{2}\right) = \frac{1}{2} \left(\frac{1}{3} mL^2\right) \omega^2 \]

### Step 7: Simplify and Solve for Angular Velocity
Cancelling \( m \) from both sides and simplifying gives:
\[ g\left(\frac{L}{2}\right) = \frac{1}{6} L^2 \omega^2 \]
\[ \omega^2 = \frac{3g}{L} \]

### Step 8: Find Linear Velocity
Substituting \( \omega \) back into the equation for linear velocity:
\[ v = \omega L = L \sqrt{\frac{3g}{L}} = \sqrt{3gL} \]

### Final Answer
The linear velocity of the top of the rod when it hits the floor is:
\[ v = \sqrt{3gL} \]

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