A thin uniform rod of mass m and length l is hinged at the lower end to a level floor and stands vertically. It is now allowed to fall, then its upper end will strike the floor with a velocity given by

Options

A. `sqrt(2gl)`
B. `sqrt(3gl)`
C. `sqrt(5gl)`
D. `sqrt(mgl)`

Explanation

To solve the problem of a thin uniform rod of mass \( m \) and length \( l \) that is hinged at one end and allowed to fall, we will use the principle of conservation of energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial and Final States
Initially, the rod is vertical, and all its potential energy is due to its height. When it falls, it rotates about the hinge, and when it strikes the floor, its potential energy is converted into rotational kinetic energy.

### Step 2: Calculate the Initial Potential Energy
The center of mass of the rod is at a height of \( \frac{l}{2} \) from the ground. The potential energy (PE) at this position is given by:
\[ PE_{\text{initial}} = mgh = mg\left(\frac{l}{2}\right) = \frac{mgl}{2} \]

### Step 3: Calculate the Final Kinetic Energy
When the rod strikes the floor, all the potential energy has been converted into rotational kinetic energy (KE). The rotational kinetic energy of a rigid body is given by:
\[ KE_{\text{final}} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the rod about the hinge and \( \omega \) is the angular velocity.

### Step 4: Find the Moment of Inertia
The moment of inertia \( I \) of a thin rod about one end is given by:
\[ I = \frac{1}{3} m l^2 \]

### Step 5: Relate Angular Velocity to Linear Velocity
The relationship between the linear velocity \( v \) of the upper end of the rod and the angular velocity \( \omega \) is given by:
\[ v = \omega l \quad \Rightarrow \quad \omega = \frac{v}{l} \]

### Step 6: Set Up the Energy Conservation Equation
Using conservation of energy:
\[ PE_{\text{initial}} = KE_{\text{final}} \]
Substituting the expressions we derived:
\[ \frac{mgl}{2} = \frac{1}{2} \left(\frac{1}{3} m l^2\right) \left(\frac{v}{l}\right)^2 \]

### Step 7: Simplify the Equation
Cancelling \( m \) from both sides and simplifying:
\[ \frac{gl}{2} = \frac{1}{6} l v^2 \]
Multiplying both sides by 6:
\[ 3gl = lv^2 \]
Dividing both sides by \( l \):
\[ 3g = v^2 \]

### Step 8: Solve for Velocity
Taking the square root of both sides gives:
\[ v = \sqrt{3g} \]

### Conclusion
The velocity of the upper end of the rod when it strikes the floor is:
\[ v = \sqrt{3g} \]