Question

A body of mass m is at height h. When it reaches the ground, it has pure rotational kinetic energy. Find its angular speed at ground. (given that moment of inertia of body is I)
1. \(​ \omega = \sqrt{mgh \over I}\)
2. \(​ \omega = \sqrt{2mgh \over I}\)
3. \(​ \omega = \sqrt{mgh \over2 I}\)
4. \(\omega = \sqrt { mgh I }\)

Answer 1

Correct Answer - Option 2 : \(​ \omega = \sqrt{2mgh \over I}\)

CONCEPT: 

• Moment of Inertia: A quantity expressing a body's tendency to resist angular acceleration, that is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation, is called the moment of Inertia.
• Rotational energy or angular kinetic energy: The kinetic energy in a body due to the rotation of it. Mathematically expressed by:

​E = 1/2 (I × ω2) 

where E is the rotational energy, I is the moment of Inertia and ω is angular velocity.

• The gravitational potential energy at height h is given by 

​​E = m g h

where m is mass of a body, g is the gravitational acceleration, and h is the height.

EXPLANATION:

When at height 'h', the body will have gravitational potential energy = mgh

• When it comes to the ground level it will have rotational kinetic energy only.
• Here, the gravitational potential energy has been converted into rotational kinetic energy.
• The rotational kinetic energy of the body is equal to the gravitational potential energy.

\(\frac{1}{2}× I × \omega^2 =mgh\)

\( \omega = \sqrt{2mgh \over I}\)

So the correct answer is option 2.