A thin bar of mass `m` and length `l` is free to rotate about a fixed horizontal axis through a point at its end.The bar is brought to a horizontal position. `(theta = 90^(@))` and then released. The angular velocity when it reaches the lowest point is

Options

A. (a)directly proportional to its length and inversely proportional to its mass
B. (b)independent of mass and inversely proportional to the square root of its length
C. (c)dependent only upon the acceleration due to gravity and the mass of the bar
D. (d)directly proportional to its length and inversely proportional to its length and inversely proportional to the acceleration due to gravity.

Explanation

To solve the problem of finding the angular velocity of a thin bar of mass `m` and length `l` when it reaches the lowest point after being released from a horizontal position, we can follow these steps:

### Step 1: Understand the Energy Conservation Principle
When the bar is released from the horizontal position, it converts its potential energy into rotational kinetic energy as it swings down to the vertical position.

### Step 2: Calculate the Initial Potential Energy
The initial potential energy (PE_initial) of the bar when it is horizontal can be calculated using the formula:
\[ PE_{\text{initial}} = mgh \]
where \( h \) is the height of the center of mass above the lowest point. Since the center of mass of the bar is at a distance of \( \frac{l}{2} \) from the pivot point, we have:
\[ h = \frac{l}{2} \]
Thus,
\[ PE_{\text{initial}} = mg \left(\frac{l}{2}\right) = \frac{mgl}{2} \]

### Step 3: Calculate the Final Rotational Kinetic Energy
The rotational kinetic energy (KE_final) of the bar when it reaches the lowest point is given by:
\[ KE_{\text{final}} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the bar about the pivot point (end of the bar). The moment of inertia of a thin rod about an axis through one end is:
\[ I = \frac{1}{3} ml^2 \]
Therefore,
\[ KE_{\text{final}} = \frac{1}{2} \left(\frac{1}{3} ml^2\right) \omega^2 = \frac{ml^2 \omega^2}{6} \]

### Step 4: Apply Conservation of Energy
According to the conservation of energy, the initial potential energy is equal to the final kinetic energy:
\[ PE_{\text{initial}} = KE_{\text{final}} \]
Substituting the expressions we found:
\[ \frac{mgl}{2} = \frac{ml^2 \omega^2}{6} \]

### Step 5: Simplify and Solve for Angular Velocity
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{gl}{2} = \frac{l^2 \omega^2}{6} \]
Now, multiply both sides by 6:
\[ 3gl = l^2 \omega^2 \]
Dividing both sides by \( l^2 \):
\[ \omega^2 = \frac{3g}{l} \]
Taking the square root of both sides gives:
\[ \omega = \sqrt{\frac{3g}{l}} \]

### Final Answer
Thus, the angular velocity \( \omega \) when the bar reaches the lowest point is:
\[ \omega = \sqrt{\frac{3g}{l}} \]

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