A bar of length l carrying a small mass m at one of its ends rotates with a uniform angular speed `omega` in a vertical plane about the mid-point of the bar . During the rotation, at some instant of time when the bar is horizontal, the mass is detached from the bar but the bar continues to rotate with same `omega` . The mass moves vertically up, comes back and reaches the bar at the same point. At that place, the acceleration due to gravity is g

Options

A. This is possible if the quantity `(omega^(2)l)/(2pig)` is an integer
B. The total time of flight of the mass is proportional to `omega^(2)`
C. The total distance travelled by the mass in air is proportional to `omega_(2)`
D. The total distance travelled by the mass in air and its total time of flight are both independent on its mass

Explanation

To solve the problem, we need to analyze the motion of the mass \( m \) when it detaches from the bar and how it relates to the rotation of the bar. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
The bar of length \( l \) is rotating about its midpoint with an angular speed \( \omega \). When the bar is horizontal, the mass \( m \) is at one end of the bar. The distance from the midpoint to the end of the bar is \( \frac{l}{2} \).

### Step 2: Determine the Linear Velocity of the Mass
The linear velocity \( v \) of the mass \( m \) at the moment it detaches can be calculated using the relationship between linear velocity and angular velocity:
\[ v = \omega \cdot r \]
where \( r = \frac{l}{2} \) (the distance from the pivot to the mass). Thus,
\[ v = \omega \cdot \frac{l}{2} \]

### Step 3: Analyze the Motion of the Mass After Detachment
Once the mass detaches, it moves vertically upwards under the influence of gravity. The time of flight \( T \) until it returns to the same vertical level can be calculated using the kinematic equations. The mass will first rise to a maximum height and then fall back down.

### Step 4: Calculate the Time of Flight
The time to reach the maximum height \( h \) can be found using:
\[ h = \frac{v^2}{2g} \]
Substituting \( v = \omega \cdot \frac{l}{2} \):
\[ h = \frac{(\omega \cdot \frac{l}{2})^2}{2g} = \frac{\omega^2 l^2}{8g} \]
The time to reach this height is given by:
\[ t_{up} = \frac{v}{g} = \frac{\omega \cdot \frac{l}{2}}{g} = \frac{\omega l}{2g} \]
The total time of flight \( T \) (up and down) is:
\[ T = 2t_{up} = 2 \cdot \frac{\omega l}{2g} = \frac{\omega l}{g} \]

### Step 5: Determine the Number of Rotations During Time \( T \)
While the mass is in the air, the bar continues to rotate. The number of rotations \( N \) made by the bar in time \( T \) can be calculated as:
\[ N = \frac{\text{Total angle rotated}}{2\pi} = \frac{\omega T}{2\pi} \]
Substituting \( T = \frac{\omega l}{g} \):
\[ N = \frac{\omega \cdot \frac{\omega l}{g}}{2\pi} = \frac{\omega^2 l}{2\pi g} \]

### Step 6: Analyze the Proportionality and Independence
From the calculations:
- The total time of flight \( T \) is proportional to \( \omega \).
- The distance traveled by the mass in the air is proportional to \( \omega^2 \).
- Both the time of flight and distance traveled are independent of the mass \( m \).

### Conclusion
Thus, the correct statements derived from the analysis are:
1. The total time of flight is proportional to \( \omega \).
2. The total distance traveled by the mass in the air is proportional to \( \omega^2 \).
3. The total time of flight and distance traveled are independent of the mass \( m \).

### Final Answer
The correct options are A, C, and D.

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