Rotational energy is nothing but the energy possessed by a rotating body. This can also be termed as rotational kinetic energy. Because, here energy is due to the rotational motion of the object. The concept of rotation energy is similar to the linear energy. Only the difference is mass and velocity is replaced by moment of inertia and angular momentum.

The rotational kinetic energy is not a new kind of kinetic energy.It is simply the sum of the ordinary translational kinetic energies of all the particles of the body. Even though the entire body may not be in translational motion, each particle has a tangential velocity and thus each particle has a kinetic energy. The instantaneous direction of each particle's velocity change as the body rotates, but kinetic energy depends on v^{2} and is a scalar, so there is no direction associated with it. It is therefore quite proper to add the kinetic energies of the particles of the rotating body.
The rotational kinetic energy is merely a convenient way of expressing the total kinetic energy of all the particles in the rigid body. The formula for rotational kinetic energy is as follows.
$\omega$ is the angular velocity
Rotational kinetic energy is similar to the kinetic energy, but replace m and v by I and $\omega$.
K = $\frac{1}{2}$I$\omega^{2}$
Where I is the moment of inertia $\omega$ is the angular velocity
Rotational kinetic energy is similar to the kinetic energy, but replace m and v by I and $\omega$.
The conservation of energy principle is true, of course for the rotational motion of a solid object. The analogies with the linear case are so close that we will just state the expressions for the various forms of energy. If you note the similarities between the rotational and linear expressions, it will make them easier to remember. This conservation law is helps to analyze the motion of systems involving objects that can rotate about an axis as well as move translationally. There is no separate conservation law for rotational motion; instead, the kinetic energy may contain both rotational and translational terms.
One of the important source of energy is earth's rotational kinetic energy. Rotational kinetic energy can be calculated using the formula,
k = $\frac{1}{2}$I$\omega^{2}$
Moment of inertia of earth is 8.070 $\times10^{37}$ kgm^{2} and angular velocity is $\frac{6.2832}{86164.09}$ radians/sec. So the rotational energy become,K = $\frac{1}{2}$$\times8.070 \times10^{37}\times(\frac{6.2832}{86164.09})^{2} = 2.138 $\times10^{29}$ J
Rotational kinetic energy can be calculated using the above formula. Let us discuss some of the problems related to the topic.
Solved Examples
Question 1: Calculate the kinetic energy of the earth's rotation about its axis?
Solution:
We know that,
Mass of the earth, M = 6.0$\times10^{24}$ kg; Radius of the earth, R = 6.4$\times10^{6}$ m
Moment of inertia of the earth is given by,
I = $\frac{2}{5}$M$R^{2}$ = $\frac{2}{5}$$\times6.0\times10^{24}\times(6.4\times10^{6})^{2}$ = 9.8$\times10^{37}$kgm^{2}
Angular velocity of the earth is 2$\pi$ radians/day
So, $\frac{2\pi}{3600\times24}$ = 7.3$\times10^{5}$rad/s
Rotational kinetic energy is given by,
K = $\frac{1}{2}$I$\omega^{2}$
K = $\frac{1}{2}$$\times9.8\times10^{37}\times(7.3\times10^{5})^{2}$ = 2.6$\times10^{29}$ J
Solution:
We know that,
Mass of the earth, M = 6.0$\times10^{24}$ kg; Radius of the earth, R = 6.4$\times10^{6}$ m
Moment of inertia of the earth is given by,
I = $\frac{2}{5}$M$R^{2}$ = $\frac{2}{5}$$\times6.0\times10^{24}\times(6.4\times10^{6})^{2}$ = 9.8$\times10^{37}$kgm^{2}
Angular velocity of the earth is 2$\pi$ radians/day
So, $\frac{2\pi}{3600\times24}$ = 7.3$\times10^{5}$rad/s
Rotational kinetic energy is given by,
K = $\frac{1}{2}$I$\omega^{2}$
K = $\frac{1}{2}$$\times9.8\times10^{37}\times(7.3\times10^{5})^{2}$ = 2.6$\times10^{29}$ J
Question 2: Calculate the rotational kinetic energy of a wheel, whose radius is 0.4m and mass 2kg and rotate with a velocity of 10 rad/s?
Solution:
From the question, it is given that,
Mass of the wheel, M = 2kg; Radius of the wheel, R = 0.4 m and angular velocity, $\omega$ = 10 rad/s
Moment of inertia of the wheel is given by,
I = $\frac{1}{2}$M$R^{2}$ = $\frac{1}{2}$$\times2\times(0.4)^{2}$ = 0.16kgm^{2}
Rotational kinetic energy is given by,
K = $\frac{1}{2}$I$\omega^{2}$
K = $\frac{1}{2}$$\times0.16\times(10)^{2}$ = 8 J
Solution:
From the question, it is given that,
Mass of the wheel, M = 2kg; Radius of the wheel, R = 0.4 m and angular velocity, $\omega$ = 10 rad/s
Moment of inertia of the wheel is given by,
I = $\frac{1}{2}$M$R^{2}$ = $\frac{1}{2}$$\times2\times(0.4)^{2}$ = 0.16kgm^{2}
Rotational kinetic energy is given by,
K = $\frac{1}{2}$I$\omega^{2}$
K = $\frac{1}{2}$$\times0.16\times(10)^{2}$ = 8 J