## Table of Contents

- What is Rotational Kinetic Energy?
- Relationship Between Angular Momentum and Kinetic Energy
- Total Kinetic Energy Formula for a Rotating Body
- Example: How to Find Rotational Kinetic Energy
- Lesson Summary

This lesson covers rotational energy. It describes the differences between rotational kinetic energy and translational kinetic energy, and it explains how to find rotational kinetic energy.
Updated: 01/05/2022

- What is Rotational Kinetic Energy?
- Relationship Between Angular Momentum and Kinetic Energy
- Total Kinetic Energy Formula for a Rotating Body
- Example: How to Find Rotational Kinetic Energy
- Lesson Summary

Kinetic energy is the energy of motion. Any moving object has kinetic energy. For an object with a linear trajectory, such as a car driving straight through the flat lands of Kansas, the equation for kinetic energy is {eq}KE_{linear} = \frac{1}{2}mv^2 {/eq} where m is mass and v is velocity. How can the kinetic energy of that same car be described once it begins driving over the foothills of the Rockies as it reaches Colorado? Driving over a hill, like a satellite orbiting the earth or a ball spinning, has a circular trajectory. The concept of **rotational kinetic energy** or **angular kinetic energy** is used describe the motion of an object with a non-linear trajectory. The formula used to describe the kinetic energy of objects that have a circular or curved trajectory will be discussed in the next section. The kinetic energy associated with the spin of the object is measured in Joules (J).

The **rotational kinetic energy formula ** is {eq}KE_{rotational} = \frac{1}{2} I \omega^2 {/eq}. The squared term is the lowercase Greek letter omega. In physics, this is the symbol used for **angular velocity**. Angular velocity is defined as the change in angle, {eq}\theta {/eq}, divided by the change in time, and mathematically it is written {eq}\omega = \frac{ \Delta \theta}{ \Delta t} {/eq}. Angular velocity is measured in radians per second.

The rotational kinetic energy formula also contains the variable I. In physics, I stands for the **moment of inertia** of an object. The moment of inertia of a rigid body is a measure of the object's mass multiplied by the perpendicular distance from the center of mass to the rotational axis of the object. The moment of inertia is a self-contained point that an object rotates around, and this measurement depends on the object's geometry. Calculations of the moment of inertia are common in integral calculus, and Figure 1 lists the general formula for calculating the moment of inertia of some common shapes.

The rotational energy formula, {eq}KE_{rotational} = \frac{1}{2} I \omega^2 {/eq}, used to find the kinetic energy of a rotating object such as a spinning electron, looks very similar to the linear kinetic energy formula, {eq}KE_{linear} = \frac{1}{2}mv^2 {/eq}, which is used to find the **translational kinetic energy** of an object such as a ball dropping from a balcony. While these formulas have a similar form, they each use different quantities. The first difference to note is that translational kinetic energy requires finding the linear velocity, v, of the moving object while the rotational kinetic energy formula requires finding the angular velocity, {eq}\omega {/eq}. The second important difference to note is that rotational kinetic energy requires calculating the moment of inertia, I, while translational kinetic energy only requires knowing the mass, m, of the moving object. Why is this? Another way to conceptualize mass is as the ability of an object to resist linear motion. An object's moment of inertia is the rotational analogue of this and is the ability of an object to resist circular motion.

To fully describe the movement of any moving object, use the **total kinetic energy formula**, {eq}KE_{total} = \frac{1}{2} (mv^2 + I \omega^2) {/eq}. In the total kinetic energy formula, if the object has only linear motion, the first term will be non-zero and the second term will be zero, whereas if the object has only rotational motion, the second term will be non-zero and the first term will be zero.

**Angular momentum**, L, is given by the formula {eq}L = mvr {/eq}. It is the rotational equivalent of linear momentum and describes the relationship between an object's mass, velocity, and position. The linear velocity, v, relates to angular velocity through the equation {eq}v = r \omega {/eq}. This can be substituted into the angular momentum equation to give {eq}L = m \omega r^2 {/eq}. Now look back to Figure 1 and notice that every moment of inertia equation involves {eq}mr^2 {/eq}. Using {eq}I = mr^2 {/eq} and substituting I into the angular momentum equation gives {eq}L = I \omega {/eq}. Hopefully this is beginning to look familiar. Remember that rotational kinetic energy has the equation {eq}KE_{rotational} = \frac{1}{2} I \omega^2 {/eq}. L can be submitted in to give {eq}KE_{rotational} = \frac{1}{2} L \omega {/eq}. Using angular momentum instead of angular velocity is another way to describe rotational kinetic energy.

Think back to the example of car driving through the Rocky Mountains. The previous examples looked at the motion of the car as a whole to decide if the trajectory was linear or circular, but what if, instead of using the whole car, only the tire was used? What kind of motion does a moving tire represent? Imagine marking a spot on a tire and then sending it rolling down the street. The spot on the tire will move in a circular motion as the tire travels in a line from its starting position to its ending position somewhere down the road. Because the tire moves in a circular motion, the kinetic energy formula will have a non-zero rotational component, {eq}KE_{rotational} = \frac{1}{2} I \omega^2 {/eq}. Referring to Figure 1, the tire can be approximated as a hollow loop so the moment of inertia of the tire is {eq}I = mr^2 {/eq} where m is the mass of the tire and r is its radius. Substituting this into the rotational kinetic energy formula, the formula becomes {eq}KE_{rotational} = \frac{1}{2} mr^2 \omega^2 {/eq}

To fully describe the motion of this tire, the linear displacement from the starting and ending positions of the tire must also be taken into account. This requires a non-zero translational kinetic energy as well, {eq}KE_{linear} = \frac{1}{2}mv^2 {/eq}. The linear velocity, v, relates to angular velocity through the equation {eq}v = r \omega {/eq}, and {eq}r \omega {/eq} can be used to replace v in the translational energy equation to give {eq}KE_{linear} = \frac{1}{2}m( r \omega)^2 {/eq}. In this example both terms of the total kinetic energy formula are non-zero. Thus, putting it all together, {eq}KE_{tire-total} = \frac{1}{2} (m( r \omega)^2 + mr^2 \omega^2) {/eq}.

Example 1: Calculate the total kinetic energy of a beach ball of mass 0.1 kg and radius 0.3 meters rolling downhill at a speed of 10 m/s.

The total kinetic energy formula is {eq}KE_{total} = \frac{1}{2} (mv^2 + I \omega^2) {/eq}.

1) State the knowns of the problem.

- r = 0.3 m
- v = 10 m/s
- m = 0.1 kg
- a beach ball is hollow, and it can be approximated as a hollow spherical shell. Using Figure 1, {eq}I = \frac{2}{3}mr^2 {/eq}.

2) All of the variables for the total kinetic energy are given except for the angular velocity of the beach ball. The linear velocity, {eq}v = r \omega {/eq}, can be rewritten in terms of angular velocity: {eq}\omega = \frac{r}{v} {/eq}. Plugging in r and v from 1) gives:

{eq}\omega = \frac{0.3 m}{10 m/s} {/eq}

3) Plugging the results from 1) and 2) into the total kinetic energy equation gives:

{eq}KE_{total} = \frac{1}{2} (0.1(10)^2 + \frac{2}{3}mr^2 (\frac{0.3}{10})^2) {/eq}

4) The final step is to replace the variables in the moment of inertia equation with the appropriate values and simplify.

{eq}KE_{total} = \frac{1}{2} (0.1(10)^2 + \frac{2}{3} 0.1*0.3^2 (\frac{0.3}{10})^2) {/eq}

= {eq}KE_{total} = 5.0000027 {/eq} J

This rounds to 5 J. Thus, the rolling beach ball has 5 Joules of kinetic energy.

When calculating the rotational kinetic energy of an object, make sure to properly identify how to model the object according to Figure 1. To find the angular velocity required to calculate the moment of inertia from Figure 1, use the information given in the problem, converting linear velocity to angular velocity if necessary. The following example shows **how to find the rotational kinetic energy** of a solid cylinder.

Example 2: Find the rotational kinetic energy associated with a tin can of diameter 8 cm that is full of Spam and weighs 0.01 kg if the can is being kicked between two children. As the can spins, it makes one revolution in 0.2 seconds.

1) This problem asks about rotational kinetic energy, so the only formula needed is {eq}KE_{rotational} = \frac{1}{2} I \omega^2 {/eq}.

2) Write the givens and make the necessary conversions.

- d = 8 cm = 0.08 m -> {eq}\frac{d}{2} = r {/eq} so {eq}r = \frac{.08}{2} = 0.04 m {/eq}

- m = 0.1 kg
- The tin can full of Spam can be modeled from Figure 1 as a solid cylinder. The moment of inertia is therefore {eq}\frac{1}{2} mr^2 {/eq}.

2) The can makes 1 revolution in 0.2 s. Angular velocity is the change in angle over the change in time. Here, the change in angle is {eq}2 \pi {/eq} radians. This change happens in 0.2 s so the angular velocity is {eq}\omega = \frac{2 \pi}{0.2} = 31.41 {/eq} rad/sec.

3) The final piece of the puzzle is to calculate the moment of inertia, {eq}\frac{1}{2} mr^2 {/eq}. The mass was given in the problem as 0.01 kg. The radius was calculated in 1) as 0.04 m. Plugging these values into the moment of inertia equation and simplifying gives:

{eq}I = \frac{1}{2} (0.01*0.04^2) = 8 * 10^{-6} {/eq} {eq}kg/m^2 {/eq}.

4) Using the values from 1) - 3) in the rotational kinetic energy formula gives:

{eq}KE_{rotational} = \frac{1}{2}(8 * 10^{-6} * 31.41 ^2) = 0.0039 {/eq} J.

**Rotational kinetic energy**, also called **angular kinetic energy**, is the energy associated with a circular trajectory. It is the energy of spinning and the **rotational kinetic energy formula ** requires knowing the **angular velocity** and the **moment of inertia** of the spinning object. The **rotational kinetic energy equation ** can also be calculated using **angular momentum**.

Many motions have a both circular and a linear component. When this is the case, the total kinetic energy must include contributions from the rotational kinetic energy and the **translational kinetic energy**. In general, **how to find the rotational kinetic energy** of an object depends on its shape and its speed.

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Frequently Asked Questions

To calculate rotational kinetic energy, both the moment of inertia of the object in motion and its angular velocity must be known. Angular velocity can be calculated from linear velocity with the equation w = r/v where w is the angular velocity, r is the radius, and v is the velocity. The moment of inertia can be looked up.

Any object that moves has kinetic energy, and any object that moves in a circular trajectory has rotational kinetic motion. A rolling tire, a spinning electron, and a car driving over a hill are all examples of motion that has rotational kinetic energy.

In physics, energy is measured in units of Joules. Rotational kinetic energy follows the same convention and is calculated in units of Joules.

Any object that moves has kinetic energy. If an object moves in a circular motion is has rotational kinetic energy that is given by the formula KE_rotational = Iw^2 where I is the moment of inertia of the object, and w is the angular velocity of the object.

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