there is a uniform bar of mass m and length l hinged at the centre so that it can turn i in a vertical plane about a horizantal axis.two masses m and 2m are rigidly attached to the end of arrangments is relased from rest them find initial angular acc of bar and 2m mass.2 intial reaction at hinge,angular speed aquired by bar if it becom vertical..

To analyze the problem of a uniform bar hinged at its center with two masses attached at its ends, we need to apply concepts from rotational dynamics and energy conservation. Let's break this down step by step.

Understanding the System

We have a uniform bar of mass \( m \) and length \( l \) that is hinged at its center. Two masses, \( m \) and \( 2m \), are attached to the ends of the bar. When the system is released from rest, we want to find the initial angular acceleration of the bar and the mass \( 2m \), the initial reaction at the hinge, and the angular speed acquired by the bar when it becomes vertical.

Finding the Initial Angular Acceleration

To find the initial angular acceleration (\( \alpha \)), we can use Newton's second law for rotation, which states:

• \( \tau = I \alpha \)

Where \( \tau \) is the net torque about the hinge, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration.

Calculating the Moment of Inertia

The moment of inertia \( I \) of the bar about the hinge is given by:

• \( I_{\text{bar}} = \frac{1}{12} m l^2 + m \left(\frac{l}{2}\right)^2 + 2m \left(\frac{l}{2}\right)^2 \)

Here, the first term is for the bar itself, and the other two terms account for the masses \( m \) and \( 2m \) located at a distance \( \frac{l}{2} \) from the hinge. Simplifying this gives:

• \( I = \frac{1}{12} m l^2 + m \frac{l^2}{4} + 2m \frac{l^2}{4} = \frac{1}{12} m l^2 + \frac{3m l^2}{4} = \frac{10m l^2}{12} = \frac{5m l^2}{6} \)

Calculating the Torque

The torque due to the gravitational forces acting on the masses is:

• \( \tau = r \cdot F \)

For mass \( m \), the torque is:

• \( \tau_m = \frac{l}{2} \cdot mg \)

For mass \( 2m \), the torque is:

• \( \tau_{2m} = \frac{l}{2} \cdot 2mg \)

The total torque is:

• \( \tau_{\text{total}} = \frac{l}{2} mg + \frac{l}{2} \cdot 2mg = \frac{l}{2} \cdot 3mg = \frac{3lmg}{2} \)

Finding Angular Acceleration

Now we can substitute the values into the equation for angular acceleration:

• \( \alpha = \frac{\tau}{I} = \frac{\frac{3lmg}{2}}{\frac{5ml^2}{6}} = \frac{3g}{2} \cdot \frac{6}{5l} = \frac{18g}{10l} = \frac{9g}{5l} \)

Initial Reaction at the Hinge

The reaction force at the hinge can be found by considering the forces acting on the system. The vertical forces include the weights of the two masses:

• Total weight = \( mg + 2mg = 3mg \)

The hinge must provide a reaction force equal to the total weight to maintain equilibrium in the vertical direction:

• Reaction force \( R = 3mg \)

Angular Speed When the Bar Becomes Vertical

To find the angular speed (\( \omega \)) when the bar becomes vertical, we can use the conservation of energy principle. The potential energy lost by the masses will convert into rotational kinetic energy:

• Potential energy lost = \( mgh + 2mgh = 3mgh \)
• Kinetic energy = \( \frac{1}{2} I \omega^2 \)

Setting the potential energy equal to the kinetic energy gives:

• \( 3mgh = \frac{1}{2} \cdot \frac{5ml^2}{6} \cdot \omega^2 \)

Solving for \( \omega \):

• \( \omega^2 = \frac{36gh}{5l^2} \)
• \( \omega = \sqrt{\frac{36gh}{5l^2}} = \frac{6\sqrt{gh}}{\sqrt{5}l} \)

In summary, the initial angular acceleration of the bar is \( \frac{9g}{5l} \), the initial reaction at the hinge is \( 3mg \), and the angular speed acquired by the bar when it becomes vertical is \( \frac{6\sqrt{gh}}{\sqrt{5}l} \). This analysis combines principles of rotational dynamics and energy conservation to provide a comprehensive understanding of the system's behavior.