# Rotational Kinetic Energy: Definition, Formula & Units (w/ Examples)

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Rotational kinetic energy​ describes the energy of motion resulting from an object’s rotation or circular motion. Recall that ​linear kinetic energy​ of a mass ​m​ moving with speed ​v​ is given by 1/2mv2. This is a straightforward calculation for any object moving in a straight-line path. It applies to the object’s center of mass, allowing the object to be approximated as a point mass.

Now, if we want to describe the kinetic energy of an extended object undergoing more complex motion, the calculation becomes trickier.

We could make successive approximations by breaking up the extended object into small pieces, each of which can be approximated as a point mass, and then compute the linear kinetic energy for each point mass separately, and add them all up to find the total for the object. The smaller we break the object up, the better the approximation. In the limit where the pieces become infinitesimal, this can be done with calculus.

But we’re in luck! When it comes to rotational motion, there is a simplification. For a rotating object, if we describe its mass distribution about the axis of rotation in terms of its moment of inertia, ​I​, we are then able to use a simple rotational kinetic energy equation, discussed later in this article.

## Moment of Inertia

Moment of inertia​ is a measure of how difficult it is to cause an object to change its rotational motion about a particular axis. The moment of inertia for a rotating object depends not only on the mass of the object, but also how that mass is distributed about the axis of rotation. The further away from the axis that the mass is distributed, the harder it is to change its rotational motion, and hence the greater the moment of inertia.

The SI units for moment of inertia are kgm2 (which is consistent with our notion that it depends on mass and on distance from rotational axis). The moments of inertia for different objects can be found in a table or from calculus.

#### Tips

• The moment of inertia for any object can be found using calculus and the formula for the moment of inertia of a point mass.

## Rotational Kinetic Energy Equation

The formula for rotational kinetic energy is given by:

KE_{rot} = \frac{1}{2}I\omega^2

Where ​I​ is the object's moment of inertia and ​ω​ is the object's angular velocity in radians per second (rad/s). The SI unit for rotational kinetic energy is the joule (J).

The form of the rotational kinetic energy formula is analogous to the translational kinetic energy equation; moment of inertia plays the role of mass, and angular velocity replaces linear velocity. Note that the rotational kinetic energy equation gives the same result for a point mass as the linear equation does.

If we imagine a point mass ​m​ moving in a circle of radius ​r​ with speed ​v​, then its angular velocity is ω = v/r and its moment of inertia is mr2. Both kinetic energy equations give the same result, as expected:

KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(mr^2)(v/r)^2=\frac{1}{2}\frac{m\cancel{r^2}v^2}{\cancel{r^2}} = \frac{1}{2}mv^2 = KE_{lin}

If an object is both rotating and its center of mass is moving along a straight line path (as happens with a rolling tire, for example), then the ​total kinetic energy​ is the sum of the rotational kinetic energy and the translational kinetic energies:

KE_{tot} = KE_{rot}+KE_{lin} = \frac{1}{2}I\omega^2+\frac{1}{2}mv^2

## Examples Using the Rotational Kinetic Energy Formula

The rotational kinetic energy formula has many applications. It can be used to calculate the simple kinetic energy of a spinning object, to calculate the kinetic energy of a rolling object (an object undergoing both rotational and translational motion) and to solve for other unknowns. Consider the following three examples:

Example 1:​ Earth spins about its axis approximately once every 24 hours. If we assume it has a uniform density, what is its rotational kinetic energy? (The radius of the earth is 6.37 × 106 m, and its mass is 5.97 × 1024 kg.)

To find the rotational kinetic energy, we first must find the moment of inertia. By approximating Earth as a solid sphere, we get:

I = \frac{2}{5}mr^2 = \frac{2}{5}(5.97\times10^{24}\text{ kg})(6.37\times10^6\text{ m})^2 = 9.69\times10^{37}\text{ kgm}^2

So the rotational kinetic energy of the Earth is then:

KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(9.69\times10^{37}\text{ kgm}^2)(7.27\times10^{-5}\text{ rad/s})^2 = 2.56\times 10^{29}\text{ J}

Fun fact: This is more than 10 times the total energy the sun puts out in a minute!

Example 2:​ A uniform cylinder of mass 0.75 kg and radius 0.1 m rolls across the floor at a constant speed of 4 m/s. What is its kinetic energy?

The total kinetic energy is given by:

KE_{tot} = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2

In this case, I = 1/2 mr2 is the moment of inertia for a solid cylinder, and ​ω​ is related to the linear velocity via ω = v/r​.

Simplifying the expression for total kinetic energy and plugging in values gives:

KE_{tot} = \frac{1}{2}(\frac{1}{2}mr^2)(v/r)^2 + \frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}mv^2 = \frac{3}{4}mv^2\\ = \frac{3}{4}(0.75\text{ kg})(4\text{ m/s}) = 2.25\text{ J}

Note that we didn't even need to use the radius! It cancelled out because of the direct relationship between rotational velocity and linear velocity.

Example 3:​ A student on a bicycle coasts down a hill from rest. If the vertical height of the hill is 30 m, how fast is the student going at the bottom of the hill? Assume the bicycle weighs 8 kg, the rider weighs 50 kg, each wheel weighs 2.2 kg (included in bicycle weight) and each wheel has a diameter of 0.7 m. Approximate the wheels as hoops and assume friction is negligible.

Here we can use mechanical energy conservation to find the final speed. The potential energy at the top of the hill is turned into kinetic energy at the bottom. That kinetic energy is the sum of the translational kinetic energy of the entire person + bike system, and the rotational kinetic energies of the tires.

Total energy of the system:

E_{tot} = PE_{top} = mgh = (50\text{ kg} + 8\text{ kg})(9.8\text{ m/s}^2)(30\text{ m}) = 17,052\text{ J}

The formula for total energy in terms of kinetic energies at the bottom of the hill is:

E_{tot} = KE_{bottom} = \frac{1}{2}I_{tires}\omega^2 + \frac{1}{2}m_{tot}v^2\\ = \frac{1}{2}(2\times m_{tire} \times r_{tire}^2)(v/r_{tire})^2 + \frac{1}{2}m_{tot}v^2\\ = m_{tire}v^2 + \frac{1}{2}m_{tot}v^2\\ = (m_{tire} + \frac{1}{2}m_{tot})v^2

Solving for ​v​ gives:

v = \sqrt{\frac{E_{tot}}{m_{tire} + \frac{1}{2}m_{tot}}}

Finally, plugging in numbers we get our answer:

v = \sqrt{\frac{17,052\text{ J}}{2.2\text{ kg} + \frac{1}{2}58\text{ kg}}} = 23.4 \text{ m/s}

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