A rod is hinged at some point away from its center and is made to oscillate like a compound pendulum. It oscillates with an angular frequency `omega`. If the rod is now rotated so that its center of mass is vertically above the point of suspension and released from this position after giving a slight push. What is its angular velocity when it reaches the lowermost position :

Options

A. `omega`
B. `2omega`
C. `omega//2`
D. `4omega`

Explanation

To solve the problem step by step, we will use the principles of conservation of energy and the properties of rotational motion.

### Step 1: Understand the Initial Setup
The rod is hinged at a point away from its center of mass. When the rod is rotated such that its center of mass is directly above the hinge and released, it will fall under the influence of gravity.

### Step 2: Determine the Initial Potential Energy
When the rod is in the initial position (let's call it point A), the center of mass is at a height of \(2d\) (where \(d\) is the distance from the hinge to the center of mass). The potential energy (PE) at this height is given by:
\[ PE_A = mgh = mg(2d) = 2mgd \]

### Step 3: Determine the Final Potential Energy
When the rod reaches the lowest position (let's call it point B), the height of the center of mass is \(0\), so the potential energy at this point is:
\[ PE_B = 0 \]

### Step 4: Apply Conservation of Energy
According to the law of conservation of energy, the total mechanical energy at point A must equal the total mechanical energy at point B. Thus:
\[ PE_A + KE_A = PE_B + KE_B \]
At point A, the kinetic energy (KE) is \(0\) because it is released from rest. Therefore:
\[ 2mgd + 0 = 0 + KE_B \]
This simplifies to:
\[ KE_B = 2mgd \]

### Step 5: Express the Kinetic Energy in Terms of Angular Velocity
The kinetic energy at point B can be expressed as rotational kinetic energy:
\[ KE_B = \frac{1}{2} I \omega_B^2 \]
where \(I\) is the moment of inertia of the rod about the hinge point and \(\omega_B\) is the angular velocity at the lowest position.

### Step 6: Set Up the Equation
Now we can set the expressions for kinetic energy equal to each other:
\[ 2mgd = \frac{1}{2} I \omega_B^2 \]

### Step 7: Solve for Angular Velocity
Rearranging the equation gives:
\[ \omega_B^2 = \frac{4mgd}{I} \]
Taking the square root of both sides, we find:
\[ \omega_B = \sqrt{\frac{4mgd}{I}} \]

### Step 8: Relate to the Given Angular Frequency
From the problem, we know that the angular frequency \(\omega\) of the oscillation is given by:
\[ \omega = \sqrt{\frac{mgd}{I}} \]
Thus, we can express \(\omega_B\) in terms of \(\omega\):
\[ \omega_B = 2\sqrt{\frac{mgd}{I}} = 2\omega \]

### Conclusion
The angular velocity of the rod when it reaches the lowermost position is:
\[ \omega_B = 2\omega \]