# Rotational Kinetic Energy Formula

## Rotational Kinetic Energy

Rotational kinetic energy is the energy absorbed by the object by the virtue of its rotation. The expressions for linear and rotational kinetic energy can be advanced in a parallel manner from the work-energy principle. Imagine the following parallel between a constant torque exerted on a flywheel with a moment of inertia I and a constant force exerted on a mass m, both beginning from rest.

For the linear case, beginning from rest, the acceleration from Newton’s second law is equivalent to the final velocity divided by the time and the average velocity is half the final velocity, presenting that the work done on the block gives it a kinetic energy equal to the work done. For the rotational case, also beginning from rest, the rotational work is τθ and the angular acceleration α provided to the flywheel is obtained from Newton’s second law for rotation. The angular acceleration is equivalent to the final angular velocity divided by the time and the average angular velocity is equal to half the final angular velocity. It follows that the rotational kinetic energy given to the flywheel is equivalent to the work done by the torque.

It is articulated by

$E_{k}=\frac{1}{2}Iw^{2}$

Where the,
Moment of inertia = I and
Angular velocity of the rotating body = ω

Rotational kinetic energy formula is made use of to calculate the rotational kinetic energy of the body in rotational motion. It is expressed in Joules (J).

Rotational kinetic Energy Problems

Here are problems on rotational kinetic energy for your reference.

Solved Examples

Problem 1: Calculate the rotational kinetic energy if angular velocity is 7.29 × 10-5 rad s-1 and moment of inertia is 8.04 × 1037kgm2.

Known:

(Angular velocity) ω = 7.29 × 10-5 rad s-1,
(Moment of inertia) I = 8.04 × 1037 kgm2

The rotational kinetic energy is $E_{k}=\frac{1}{2}Iw^{2}$

$=\frac{1}{2}\times8.04\times10^{37kgm^{2}}\times(7.29\times10^{-5}rad/s)^{2}$

= 2.13 × 1029 J

Problem 2: Determine the rotational kinetic energy of electric motor if angular velocity is 100 ππ rad/s and moment of inertia is 50 kgm2.

Angular velocity ω = 100 π rad/s

Moment of inertia I = 50 kgm2

The rotational kinetic energy is $E_{k}=\frac{1}{2}Iw^{2}$

$=\frac{1}{2}\times100\pi&space;rad/s\times50kgm^{2}$

= 2500 J

#### Practise This Question

Match the following:
1. Rate of Emission        I. Temperature of the Surroundings
2. Rate of Absorption      II. Temperature of the Body