Kinetic energy is the energy of the objects in motion. This motion may be linear or rotational as well. Thus an object in rotational motion will also possess the kinetic energy. This energy is referred to as rotational kinetic energy. The kinetic energy of a rotating object depends on the object’s angular i.e. rotational velocity expressed in radians per second. It also depends on the object’s moment of inertia. Moment of inertia is a measure of how easy it is to change the rotational state of an object. This topic will explain the rotational kinetic energy formula with examples. Let us learn it!

**Concept of Rotational Kinetic Energy**

Rotational kinetic energy is the energy absorbed by the object by the virtue of its rotation. The equations for linear and rotational kinetic energy can be expressed in the same way as to 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.

Source:pixabay.com

For the linear case, beginning from rest, the acceleration from Newtonâ€™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 kinetic energy equal to the work done. In the case of rotational motion, also beginning from rest, the rotational work will be:

**\(\tau \times \theta\)**

And the angular acceleration \(\alpha\) of the flywheel we obtain from Newtonâ€™s second law for rotation. The angular acceleration will be equivalent to the final angular velocity divided by the time. Also, the average angular velocity will be equal to half the final angular velocity. Further, the rotational kinetic energy that will be given to the flywheel is equivalent to the work done by the torque.

**The Formula for Rotational Kinetic Energy**

We can compute it with the followingÂ Rotational Kinetic Energy Formula:

\(Rotational\; kinetic\; energy = \frac{1}{2} \times (moment\; of \; inertia) \times (angular \;Â velocity)^2\)

i.e. mathematically,

\(E_{k}=\frac{1}{2}I \omega^{2}\)

\(E_{k}\) | Rotational kinetic energy |

I | Moment of inertia |

\(\omega\) | Angular velocity of the rotating body |

Rotational kinetic energy formula is useful to calculate the rotational kinetic energy of the body having rotational motion. Moments of inertia are represented with the letter I and are expressed in units of kgâˆ™m2. The unit of kinetic energy is Joules (J). In other units, one Joule will be equal to one kilogram meter squared per second squared i.e. \(kg m^2 s^{-2}.\)

**Solved Examples forÂ Rotational Kinetic Energy Formula**

Q.1:Â Calculate the rotational kinetic energy if angular velocity is\(\omega = 7.29 \times 10^{-5}rad per sec,\) and moment of inertia is\(8.04 \times 10^{37} kg m^2.\)

Solution: Known parameters are as follows,

Angular velocity, \(\omega = 7.29 \times 10^{-5}rad per sec,\)

Moment of inertia, \(I = 8.04 \times 10^{37} kg m^2.\)

The rotational kinetic energy is \(E_{k}= \frac{1}{2}I \omega ^{2}\)

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

\(= 2.13 \times 10^{29} J\)

Therefore, rotational kinetic energy will beÂ \(2.13 \times 10^{29} J.\)

Q.2: Determine the rotational kinetic energy of electric motor if angular velocity is \(100 \pi\) rad per sec and the moment of inertia is \(50 kg m^2\)

Solution:

Angular velocity, omega = \(100 \pi rad per sec\)

Moment of inertia \(I = 50 kg m^ 2\)

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

\(=\frac{1}{2}\times100 \pi rad/s\times50 kgÂ m^{2}\)

= 2500 J

Therefore, rotational kinetic energy will be 2500 J.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

Interesting studies

It is already correct f= ma by second newton formula…