How much energy do we need to rotate a wheel? How will different shapes of the object have different rotational kinetic energy? How does mass distribution affect the rotational kinetic energy? The only answer to all these questions is through the formula of rotational kinetic energy.

Rotational kinetic energy describes the energy of motion of a rotating object. Rotational kinetic energy plays the same role in rotational motion as played by translational kinetic energy in translational motion. We know that the linear kinetic energy depends on the mass and speed of the body, in a similar way, the rotational kinetic energy depends on the moment of inertia (rotational mass) and angular velocity. In this article, we will provide detailed information on the rotational kinetic energy formula. Scroll down to learn more!

Moment of Inertia

The moment of inertia of a system is a measure of how difficult it is to cause an object to change its rotational motion about a specific axis. It is the measure of rotational inertia of a system. The moment of inertia for a rotating object depends on the mass of the object, and it also depends on how the mass is distributed about the axis of rotation. If the mass is distributed further away from the axis, then the moment of inertial will be more, then it will be harder to change its rotational motion. The values of the moment of inertia for different bodies are different. The value of the moment of inertia of some of the objects of mass \(M\) and radius \(R\) about the axis of rotation passing through their centre of mass is given below.
I. MI of Circular Ring \(=\) MI of Hollow Cylinder \(= MR^2\)
II. MI of Disc \(=\) MI of Solid Cylinder \(= \frac{1}{2} MR^2\)
III. M.I Hollow Sphere \(= \frac{2}{3} MR^2\)
IV. MI Solid sphere \(= \frac{2}{5} MR^2\)
The SI unit of moment of inertia is given by \(\rm{kg}\;\rm{m}^2\) (which is consistent with our notion that it depends on mass and square of the distance from the rotational axis).

Rotational Kinetic Energy Formula

The formula of rotational kinetic energy is analogous to linear kinetic energy. We know that the linear kinetic energy of a mass \(m\) moving with speed \(v\) is given by \(\frac{1}{2}\;\rm{mv}^2\). We can assume the rigid body is made up of an infinite number of point masses. At any instant of time, each point mass will be moving with a linear velocity equal to the tangential velocity of that point. The magnitude of tangential velocity at any point at distance \(r\) from the rotating axis of a rigid body rotating with angular velocity \(ω\) is given by,
\(v = rω\)
For any point mass \(m_1\) of a rigid body, the kinetic energy will be given by,
\(K{E_1} = \frac{1}{2}{m_1}v_1^2\)
Total kinetic energy, \(K{E_R} = \sum {_1^\infty } \frac{1}{2}{m_i}v_i^2\)
\( = \sum {_1^\infty } \frac{1}{2}{m_i}r\omega _i^2\)
\( = \sum {_1^\infty } \frac{1}{2}{m_i}r_i^2{\omega ^2}\)
\( = \frac{1}{2}{\omega ^2}\sum {_1^\infty } {m_i}r_i^2\)
\( = \frac{1}{2}I{\omega ^2}\)

So, the rotational kinetic energy, \(KE_R = \frac{1}{2} Iω^2\). Here \(I\) is the rotational mass or moment of inertia of a rotating object, and \(ω\) is the angular speed.

Below we have provided the units of rotational kinetic energy in SI and MKS system:

SI unit system	Joules (J)
MKS unit system	kg.m2.s-2

Dimensional Formula of Rotational Kinetic energy

We know that rotational kinetic energy is written as, \(KE = \frac{1}{2} Iω^2.\) The moment of inertia of any object of mass \(m\) has the dimensions of \(mr^2\). Now the dimensional formula of the moment of inertia is given by, \(\left[ I \right] = {M^1}{L^2}{T^0}.\)
Also, the dimensional formula of angular velocity, \(\left[ \omega \right] = {M^0}{L^0}{T^{ – 1}}\)
By plugging the dimensional formula of the moment of inertia and angular velocity, we get:
Dimensional formula of rotational kinetic energy, \(\left[ {KE} \right] = {M^1}{L^2}{T^{ – 2}}\), which is the same as that of energy (as it should be).

Kinetic Energy of a Rigid Body in Combined Rotational and Transitional Motion

When the body has both rotational and translational motion, it has both translational kinetic energy and rotational kinetic energy.
The translational kinetic energy is given by, \(KE_T = \frac{1}{2} mv_{\rm{COM}}^2\). Here \(v_{\rm{COM}}\) is the translational velocity of the centre of mass of the body. Also, the rotational kinetic energy is given by, \(KE_R = \frac{1}{2} I_{\rm{COM}} \omega^2\). Here \(I_{\rm{COM}}\) is the moment of inertia of the rigid body about the rotating axis of the rigid body.
Thus, the total kinetic energy of the rigid body will be,
\(KE = \frac{1}{2}mv_{{\text{COM}}}^2 + \frac{1}{2}{I_{{\text{COM}}}}\omega^2\)

Rotational Kinetic Energy Formula in Terms of Angular Momentum

We know that the angular momentum of the rigid body having a moment of inertia \(I\) rotating with angular speed \(ω\) is given by, \(L = Iω\).
Also, the rotational kinetic energy of the rigid body is given by,
\(K{E_R} = \frac{1}{2}I\omega ^2\)
By multiplying the numerator and denominator of the above equation with \(I\), we will get,
\(K{E_R} = \frac{1}{2}\frac{{{I^2}\omega ^2}}{I}\)
\(K{E_R} = \frac{L^2}{2I}\)
As \(K{E_R} = \frac{L^2}{2I}\) i.e. \(K{E_R} \propto {L^2}\)
It can be represented with \(y = Kx^2\). If angular momentum is the same for two objects, then kinetic energy will be inversely proportional to the moment of inertia. So, we can say that kinetic energy will be greater for a smaller moment of inertia (If angular momentum will be the same).

Rotational Kinetic Energy Formula for a Sphere

We know that the moment of inertia of a solid sphere about an axis passing through its centre is given by,
\(I = \frac{2}{5}MR^2\) Where \(R\) is the radius of the solid sphere, and \(M\) is the mass of the solid sphere.
The rotational kinetic energy of the sphere is given by,
\(KE_R = \frac{1}{2} Iω^2\)
\(= \frac{1}{2} \times \frac {2}{5} MR^2 ω^2\)
\(= \frac{1}{5} MR^2 ω^2\)

Thus the rotational kinetic energy of a solid sphere rotating about a fixed axis passing through the centre of mass will be equal to, \(KE_R = \frac{1}{5} MR^2 ω^2\).

Rotational Kinetic Energy Formula for a Cylinder

Moment of inertia of solid cylinder about an axis passing through its cylinder axis is given by, \(I = \frac{1}{2} MR^2\) where \(R\) is the radius of the solid cylinder, and \(M\) is the mass of the solid cylinder.
The rotational kinetic energy of the solid cylinder will be,
\(KE_R = \frac{1}{2} Iω^2\)
\(= \frac{1}{4} MR^2 ω^2\)

Rotational Kinetic Energy Formula for a Disc

Moment of inertia of disc about an axis passing through its centre is given by \(I = \frac{1}{2} MR^2\) where \(R\) is the radius of the disc, and \(M\) is the mass of the disc.
The rotational kinetic energy of the disc is given by,
\(KE_R = \frac{1}{2} Iω^2\)
\( = \frac{1}{4} MR^2 ω^2\)

We can observe that the moment of inertia of a solid cylinder and disc of the same mass and same angular velocity are equal.

Solved Examples on Rotational Kinetic Energy Formula

Q.1. Earth spins about its axis approximately once every \(24\) hours. Let’s assume that it has a uniform density. What is its rotational kinetic energy? (we know that the radius of the Earth is \(6.37 × 106\,\rm{m}\), and its mass is \(5.97 × 1024\;\rm{kg}\).)
Ans: Given,
Mass of the Earth, \(m = 5.97 × 1024\;\rm{kg}\),
The radius of the Earth, \(r = 6.37 × 106\;\rm{m}\).
First, we must find the moment of inertia to find the rotational energy. By approximating Earth as a solid sphere, we get:
Moment of inertia, \(I = \frac{2}{5} mr^2\)
\(= \frac{2}{5} (5.97 × 1024\;\rm{kg}) × (6.37 × 106\;\rm{m})^2 = 9.69 × 1037\;\rm{kg}\;\rm{m}^2\)
Now the angular velocity of the Earth is \(2π\;\rm{radians}\;\rm{day}^{-1}\). Converting this to \(\rm{rad}\;\rm{s}^{-1}\) gives:
\(2\pi \;\frac{{{\text{radians}}}}{{{\text{day}}}} \times \frac{{1\;{\text{day}}}}{{86400\;{\text{seconds}}}} = 7.27 \times {10^{ – 5}}\;{\text{rad}}{{\text{s}}^{ – 1}}\)
So, the rotational \(KE\) of the Earth will be:
\(KE_R = \frac{1}{2} Iω^2\)
\(= \frac{1}{2}\left( {9.69 \times {{10}^{37}}\;{\text{kg}}\;{{\text{m}}^2}} \right) \times {\left( {7.27 \times {{10}^{ – 5}}\;{\text{rad}}\;{{\text{s}}^{ – 1}}} \right)^2}\)
\(= 2.56 × 102^9\;\rm{J}\)

Q.2. A uniform solid cylinder of mass \(0.75\;\rm{kg}\) and radius \(0.1\;\rm{m}\) rolls across the floor at a constant speed of \(4\;\rm{m}\;\rm{s}^{-1}\). What is its kinetic energy?
Ans: Given,
Mass of the solid cylinder, \(m = 0.75\;\rm{kg}\)
Radius of the cylinder, \(r = 0.1\;\rm{m}\)
Linear velocity, \(v = 4\;\rm{m}\;\rm{s}^{-1}\)
The total kinetic energy is given by:
\(KE_{\rm{tot}} = \frac{1}{2} Iω² + \frac{1}{2} mv²\)
For a uniform solid cylinder, moment of inertia is given by, \(I = \frac{1}{2} mr^2\).
Now for pure rotation without slipping, \(= \frac {v}{r}\).
Simplifying the expression for total kinetic energy and putting the all the given values, we will get:
\(K{E_{tot}} = \frac{1}{2}\left( {\frac{1}{2}m{r^2}} \right){\left( {\frac{v}{r}} \right)^2} + \frac{1}{2}m{v^2}\)
\( = \frac{1}{4}m{v^2} + \frac{1}{2}m{v^2} = \frac{3}{4}m{v^2}\)
\( = \frac{3}{4} \times 0.75 \times {4^2}\)
\(K{E_{{\text{tot}}}} = 3 \times 0.75 \times 4 = 9\;{\text{J}}\)

Summary

The kinetic energy of a rotating object is analogous to linear kinetic energy. It is given by, \(KE_R = \frac{1}{2} Iω^2\). Here \(I\) is the rotational mass or moment of inertia of a rotating object, and \(ω\) is the angular speed. The SI unit of kinetic energy is \(\rm{J}\). Its dimensional formula is \(M^1 L^2 T^{-2}\). It can also be expressed in terms of the moment of inertia and angular velocity. The rotational kinetic energy in terms of the moment of inertia and angular velocity is given by, \(KE_R = \frac{L^2}{2I}\).

The kinetic energy of the rigid body a rigid body in combined rotational and transitional motion is given by, \(KE = \frac{1}{2} mv_{\rm{COM}}^2 + \frac{1}{2} I_{\rm{COM}} ω²\). Here, rotational kinetic energy is, \(KE_R = \frac{1}{2} Iω²\), and transitional kinetic energy is, \(K_T = \frac{1}{2} mv_{\rm{COM}}^2\).

FAQs on Rotational Kinetic Energy Formula

Q.1. What are the units of rotational kinetic energy?
Ans: Unit of the rotational kinetic energy in SI and MKS system are given below:
SI unit system: Joules \((\rm{J})\)
MKS unit system: \(\rm{kg}\;\rm{m}^2\;\rm{s}^2\)

Q.2. What effect does rotation have on the total kinetic energy of a moving body?
Ans: When the body will start rotating during linear motion, its kinetic energy will increase. The total kinetic energy will be the sum of rotational kinetic energy and translational kinetic energy.

Q.3. How does rotational kinetic energy relate to linear kinetic energy?
Ans: Rotational kinetic energy is analogous to linear kinetic energy. It is given by,
\(K{E_R} = \frac{1}{2}I{\omega ^2}.\) In the case of combined rotation and translation, The kinetic energy of the rigid body a rigid body is given by, \(KE = \frac{1}{2}mv_{{\text{COM}}}^2 + \frac{1}{2}{I_{{\text{COM}}}}{\omega ^2}.\)

Q.4. Write the formula for rotational kinetic energy?
Ans: We know that rotational kinetic energy is given as \({K_{\text{R}}} = \frac{1}{2}I{\omega ^2}.\) Here \(I\) is the rotational mass or moment of inertia of a rotating object, and \(\omega\) is the angular speed.

Q.5. How is rotational kinetic energy related to angular momentum?
Ans: Rotational kinetic energy is given by, \({K_{\text{R}}} = \frac{{{L^2}}}{{2I}}\) where \(I\) is the moment of inertia, and \(L\) is the angular momentum.

Q.6. What is the SI unit of Rotational Kinetic Energy?
Ans: Joules (J) is the SI unit of rotational kinetic energy.

Q.7. Define Moment of Inertia.
Ans: The moment of inertia of a system is defined as a measure of how difficult it is to cause an object to change its rotational motion about a specific axis.

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