An uniform rod of length 2l is placed with one end in contact with the horizontal and is then inclined at an angle alpha to the horizontal and allowed to fall without slipping at contact point. When it becomes horizontal, its angular velocity will be.........

To determine the angular velocity of a uniform rod of length 2l that is inclined at an angle alpha to the horizontal and allowed to fall without slipping, we can use principles of energy conservation and rotational dynamics.

Understanding the System

The rod is initially tilted at an angle and then allowed to pivot about the contact point on the ground. As it falls, gravitational potential energy is converted into rotational kinetic energy. The final position we are interested in is when the rod becomes horizontal.

Energy Conservation Principle

We start by establishing the conservation of mechanical energy. The initial potential energy of the rod when inclined is converted into kinetic energy as it falls. The potential energy (PE) is given by:

• PE_initial = mgh

Where:

• m = mass of the rod
• g = acceleration due to gravity
• h = height of the center of mass above the horizontal position

For a uniform rod of length 2l, when it is inclined at an angle alpha, the height of its center of mass can be calculated as:

• h = l * sin(alpha)

Calculating Initial Potential Energy

Substituting the height into the potential energy formula, we have:

• PE_initial = mg(l * sin(alpha))

Rotational Kinetic Energy at Horizontal Position

When the rod becomes horizontal, all the potential energy has been converted into rotational kinetic energy (KE), which is given by:

• KE = (1/2) I ω²

Here, I is the moment of inertia of the rod about the pivot point, and ω is the angular velocity. For a rod rotating about one end, the moment of inertia is:

• I = (1/3)m(2l)² = (4/3)ml²

Setting up the Energy Equation

According to the conservation of energy:

• PE_initial = KE

Substituting the expressions we derived:

• mg(l * sin(alpha)) = (1/2) * (4/3)ml² * ω²

Simplifying the Equation

We can simplify this equation by canceling out m from both sides (assuming m ≠ 0):

• g(l * sin(alpha)) = (2/3)l² * ω²

Now, isolate ω²:

• ω² = (3g * sin(alpha)) / (2l)

Final Angular Velocity

Taking the square root of both sides gives us the angular velocity ω:

• ω = √((3g * sin(alpha)) / (2l))

This equation provides the angular velocity of the rod when it reaches the horizontal position, showcasing the balance between gravitational potential energy and rotational kinetic energy during its fall.