A thin uniform rod AB of mass m and length l is hinged at one end to the level floor and stands vertically. If it s allowed to fall, with what angular velocity will it strike the floor?

Options

A. `sqrt(mgl)`
B. `sqrt(3g/l)`
C. `sqrt(5 mgl)`
D. `sqrt(2mgl)`

Explanation

To find the angular velocity of a thin uniform rod of mass \( m \) and length \( l \) when it strikes the floor after being allowed to fall from a vertical position, we can use the principle of conservation of mechanical energy. Here's a step-by-step solution:

### Step 1: Identify the Initial Potential Energy
When the rod is in the vertical position, the center of mass of the rod is at a height of \( \frac{l}{2} \) from the floor. The initial potential energy (PE_initial) of the rod can be calculated using the formula:
\[ PE_{\text{initial}} = mgh \]
where \( h = \frac{l}{2} \). Thus,
\[ PE_{\text{initial}} = mg \left(\frac{l}{2}\right) = \frac{mgl}{2} \]

### Step 2: Identify the Final Potential Energy
When the rod falls and becomes horizontal, its center of mass is at the level of the floor, meaning its height is now 0. Therefore, the final potential energy (PE_final) is:
\[ PE_{\text{final}} = 0 \]

### Step 3: Identify the Final Kinetic Energy
When the rod is horizontal, all the initial potential energy has been converted into rotational kinetic energy (KE_final). The rotational kinetic energy 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 point. For a thin rod hinged at one end, the moment of inertia is given by:
\[ I = \frac{1}{3} ml^2 \]

### Step 4: Apply Conservation of Energy
According to the conservation of mechanical energy, the initial potential energy equals the final kinetic energy:
\[ PE_{\text{initial}} = KE_{\text{final}} \]
Substituting the values we have:
\[ \frac{mgl}{2} = \frac{1}{2} \left(\frac{1}{3} ml^2\right) \omega^2 \]

### Step 5: Simplify the Equation
We can simplify the equation:
\[ \frac{mgl}{2} = \frac{ml^2}{6} \omega^2 \]
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{gl}{2} = \frac{l^2}{6} \omega^2 \]

### Step 6: Solve for Angular Velocity \( \omega \)
Now, we can solve for \( \omega^2 \):
\[ \omega^2 = \frac{gl}{2} \cdot \frac{6}{l^2} = \frac{3g}{l} \]
Taking the square root gives us:
\[ \omega = \sqrt{\frac{3g}{l}} \]

### Final Answer
Thus, the angular velocity \( \omega \) when the rod strikes the floor is:
\[ \omega = \sqrt{\frac{3g}{l}} \]