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 plz reply imdly............. i hav to sit in test series

To solve the problem of the uniform bar hinged at its center with two masses attached, we need to analyze the forces and torques acting on the system. 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. At the ends of the bar, we have two masses: one mass \( m \) and another mass \( 2m \). 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.

Calculating Initial Angular Acceleration

To find the initial angular acceleration, we can use Newton's second law for rotation, which states that the net torque (\( \tau \)) acting on an object is equal to the moment of inertia (\( I \)) times the angular acceleration (\( \alpha \)). The equation can be expressed as:

τ = I * α

Finding the Moment of Inertia

The moment of inertia of the bar about the hinge is given by:

I_bar = (1/12) * m * l² + m * (l/2)² + 2m * (l/2)²

Here, the first term is the moment of inertia of the bar itself, and the other two terms account for the masses \( m \) and \( 2m \) located at a distance of \( l/2 \) from the hinge.

Calculating this gives:

I_bar = (1/12) * m * l² + m * (l²/4) + 2m * (l²/4)

I_bar = (1/12) * m * l² + (1/4) * m * l² + (1/2) * m * l² = (1/12 + 3/4) * m * l² = (10/12) * m * l² = (5/6) * m * l²

Calculating Torque

The torque due to the weights of the masses about the hinge can be calculated as follows:

• The torque due to mass \( m \) is \( τ_m = m * g * (l/2) \)
• The torque due to mass \( 2m \) is \( τ_{2m} = 2m * g * (l/2) \)

Thus, the total torque \( τ \) is:

τ = τ_m + τ_{2m} = m * g * (l/2) + 2m * g * (l/2) = (3m * g * l)/2

Finding Angular Acceleration

Now, substituting the values of torque and moment of inertia into the rotational equation:

(3m * g * l)/2 = (5/6) * m * l² * α

Solving for \( α \):

α = (3g)/(5l)

Initial Reaction at the Hinge

The reaction force at the hinge can be found by considering the vertical forces acting on the system. The total weight acting downwards is:

W = m * g + 2m * g = 3m * g

Since the bar is hinged and can rotate, the vertical reaction force \( R \) at the hinge must balance this weight:

R = 3m * g

Angular Speed When the Bar Becomes Vertical

To find the angular speed acquired by the bar when it becomes vertical, we can use the conservation of energy principle. The potential energy lost by the system as it falls will convert into kinetic energy:

The initial potential energy when the bar is horizontal is:

PE_initial = (m * g * (l/2)) + (2m * g * (l/2)) = (3m * g * l)/2

The kinetic energy when the bar is vertical is given by:

KE = (1/2) * I * ω²

Setting the initial potential energy equal to the kinetic energy:

(3m * g * l)/2 = (1/2) * (5/6) * m * l² * ω²

Solving for \( ω \):

ω² = (3g)/(5l)

ω = √((3g)/(5l))

In summary, the initial angular acceleration of the bar is \( (3g)/(5l) \), the initial reaction at the hinge is \( 3mg \), and the angular speed acquired by the bar when it becomes vertical is \( √((3g)/(5l)) \).