Lesson Explainer: Rotational Kinematics Physics • 9th Grade

In this explainer, we will learn how to model the change of the position with time of objects that move along circular paths.

Imagine we release a cylinder from rest at the top of an incline, as shown in the diagram below.

As the cylinder rolls down, it will experience two types of motion: linear motion down the slope and rotational motion. To visualize that rotation, say that a red dot is fixed at the center of the cylinder. The rolling cylinder moves in circlesβ€”that is, it rotatesβ€”about the red dot, as shown in the following diagram.

Now imagine we print the letter 𝑅 on the surface of the cylinder and track this letter relative to the red dot while the cylinder rolls down the slope.

At one instant in time, the letter 𝑅 will be directly above the center of the cylinder. Then, some time later, the letter will again be directly above the red dot. In the following diagram, the rolling cylinder is shown at equal distances apart, but not at equal timesβ€”the cylinder accelerates as it rolls down the incline.

Observing the letter 𝑅 directly above the red dot at two different instants in time tells us that the cylinder has gone through one complete rotation, or revolution. The cylinder has rotated through 360 degrees.

Another way to describe rotation is to use units called radians. If we take a circle with radius 𝑅 and highlight an arc of the circle’s circumference that has a length of 𝑅, then the angle of that arc length measured from the center of the circle is one radian.

A complete rotation of the circle equals an angle of exactly 2πœ‹ radians. This means that 360=2πœ‹.degreesradians

This equation shows us how to convert an angle from degrees to radians or from radians to degrees. If we divide both sides by 2πœ‹ radians, we have that: 3602πœ‹=1.degreesradians

Given an angle in radians, if we multiply it by the fraction on the left, we get that angle in degrees.

For the opposite conversion, we divide both sides of the original equation by 360 degrees, giving us 1=2πœ‹360.radiansdegrees

Given an angle in degrees, if we multiply it by the fraction on the right, we get that same angle in units of radians.

When an object rotates, we say that it has gone through an angular displacement, measured in degrees or radians.

Example 1: Converting an Angular Displacement in Radians to Degrees

Fill in the blank: An angular displacement of radians is equal to an angular displacement of 155∘. Give your answer to two decimal places.


We want to fill in the blank with the angular displacement in radians that is equal to an angular displacement of 155 degrees.

We will need to convert the angle of 155 degrees to the equivalent angle in units of radians.

To do this, we can recall that 360 degrees equals 2πœ‹ radians, which means that 2πœ‹360=1.radiansdegrees

Because this fraction equals one, we can use it to multiply 155 degrees without changing the angle’s overall value.

However, multiplying by this ratio will change the unit in which the angle is expressedβ€”from degrees to radians: 155Γ—ο€½2πœ‹360=31πœ‹36.degreesradiansdegreesradians

Rounding our result to three significant figures, we find that 155=2.71.degreesradians

Thinking back to our cylinder rolling downhill, we can say that the cylinder goes through linear as well as angular displacement. The symbol for linear displacement is 𝑠, while the symbol that represents angular displacement is πœƒ.

As the cylinder rolls along, we know it will move faster and faster down the incline. Its linear velocity increases, and we can say the same thing about the rate at which the cylinder rotates.

The cylinder’s rate of rotation is the rate at which its angular displacement changes. We call this its angular velocity, and it is completely analogous to the cylinder’s linear velocity. Angular velocity is represented by the symbol πœ”: πœ”=Ξ”πœƒΞ”π‘‘.

We can see that the rolling cylinder’s angular velocity starts out small and becomes larger over time. So, not only does the cylinder’s angular displacement change over time, but also its angular velocity does. The rate at which angular velocity changes over time is called angular acceleration. This quantity is represented by the symbol 𝛼: 𝛼=Ξ”πœ”Ξ”π‘‘.

Example 2: Calculating Angular Acceleration from Angular Rotation Rate

A drill bit is initially at rest. When the drill is activated, the drill bit rotates 47.5 times per second. The drill bit reaches this speed in a time of 175 ms. What is the angular acceleration of the drill bit? Give your answer to the nearest radian per second squared.


Here, we are told of a drill bit with an operating angular speed of 47.5 rotations every second. Starting from rest, the drill bit reaches this speed in 175 ms.

In solving for the drill bit’s angular acceleration, we would like to give our answer in units of radians per second squared rather than in revolutions per second squared.

As a first step, then, let’s convert 47.5 rotations per second to units of radians per second.

There are 2πœ‹ radians in one complete revolution. That means this angular speed equals (47.5)(2πœ‹) radians per second, or 95πœ‹ radians per second. Let’s call this speed πœ”: πœ”=95πœ‹/.rads

The drill bit increases its angular speed from 0 to πœ” over 175 ms. This means the bit goes through an angular acceleration, which in general equals the rate at which angular velocity changes with time. If we call the angular acceleration 𝛼, then 𝛼=Ξ”πœ”Ξ”π‘‘.

In this instance, Ξ”πœ” is just πœ” since the drill started at rest and Δ𝑑 is 175 ms.

We can then go ahead in calculating 𝛼: 𝛼=Ξ”πœ”Ξ”π‘‘,=95πœ‹/175.radsms

Before we solve for this fraction, we want the units of time in the numerator and the denominator to agree. We can convert 175 ms to the equivalent time in seconds. Since 1000=1,175=0.175.mssmss

Knowing this, we can now write 𝛼=95πœ‹/0.175radss and solve for 𝛼 in units of radians per second squared (rad/s2).

To three significant figures, the angular acceleration of the drill bit 𝛼 is 1β€Žβ€‰β€Ž710 rad/s2.

Note that these three quantities, angular displacement, angular velocity, and angular acceleration, all have linear analogues.

We know that linear displacement, velocity, and acceleration appear often in what are called the equations of motion. These equations describe how objects move under constant acceleration.

It turns out that we can write similar equations of motion using angular quantities. We can think of these as β€œequations of angular motion”: πœ”=πœ”+𝛼×𝑑,πœ”=πœ”+12Γ—π›ΌΓ—πœƒ,πœƒ=πœ”Γ—π‘‘+12×𝛼×𝑑,πœƒ=πœ”+πœ”2×𝑑.

These equations let us analyze and understand rotational motion just like we do linear motion.

Example 3: Computing Rotational Motion from Angular Velocity and Acceleration

The blades of a large wind turbine rotate fully in a time of 3.25 s when it is operating at its full speed. The angular acceleration of the turbine while increasing to its full speed is 0.124 rad/s2. How much time is required to bring an initially inactive turbine to its full operating speed? Give your answer to one decimal place.


In this example, we want to solve for the time needed for these turbine blades to accelerate from rest to full operating speed.

We can first note that the angular acceleration of the blades is constant. This means we can use the equations of angular motion to describe the blades’ rotation.

There are four such equations, and we will pick the one that uses information we are given and also includes the variable we want to solve for: time 𝑑.

We know the turbine’s angular acceleration, and we also know how long it takes the blades of the turbine to go through one revolution at full speed.

Recall that angular velocity is defined as a change in angular displacement divided by a change in time. Since we know the time required for the turbine to rotate once at full speed and also that one revolution equals an angular displacement of 2πœ‹ radians, we can write the maximum angular speed of the turbine: πœ”=13.25Γ—2πœ‹/.maxrevsradrev

Since the turbine starts from rest, we can consider this its final speed. This guides our choice of which equation of angular motion to work with. We can use πœ”=πœ”+π›ΌΓ—π‘‘οŒΏοƒ and rearrange it to solve for time 𝑑.

First let’s note that πœ”οƒ is zero. If we then divide both sides of the equation by the angular acceleration, we find that 𝑑=πœ”π›Ό.

We are told that 𝛼=0.124/rads, and we have calculated that πœ”=2πœ‹3.25rads, or 2πœ‹3.25 radians per second.

Therefore, 𝑑=/0.124/.οŠ¨οŽ„οŠ©οŽ–οŠ¨οŠ«οŠ¨radsrads

Units of radians per second cancel from the numerator and the denominator, and we find that 𝑑=15.5910….s

Since the values given in this example each have three significant figures, we will round our answer to that same precision. To three significant figures, the time needed for the turbine to reach its full operating speed is 15.6 seconds.

Example 4: Determining the Number of Rotations of a Rolling Object

A felled tree trunk rolls down a slope in a time of 7.2 s. The trunk is initially at rest at the top of the slope and has an angular velocity of 12 rad/s at the base of the slope. How many complete rotations does the trunk make as it rolls down the slope?


We can consider this tree trunk as a cylinder rolling down a slope as shown in the figure. As the trunk rolls, it will rotate with increasing angular speed. The rate at which its angular speed increases is constant, meaning that the trunk experiences a constant angular acceleration. Therefore, the equations of angular motion can be applied to the trunk’s motion.

These equations are as follows: πœ”=πœ”+𝛼×𝑑,πœ”=πœ”+12Γ—π›ΌΓ—πœƒ,πœƒ=πœ”Γ—π‘‘+12×𝛼×𝑑,πœƒ=πœ”+πœ”2×𝑑.οŒΏοƒοŒΏοŠ¨οƒοŠ¨οƒοŠ¨οŒΏοƒ

We want to determine the number of times the trunk rotates completely as it rolls down the slope. The number of complete revolutions an object makes correlates with its angular displacement, represented by the symbol πœƒ. If we solve for πœƒ, we will be able to convert this value to revolutions.

In this exercise, we know the trunk’s angular velocity at the start and end of its descent, and we also know how much time the descent requires. These values correspond to the variables in the fourth equation of angular motion, πœƒ=πœ”+πœ”2×𝑑.οŒΏοƒ

In our case, the initial angular velocity of the trunk, πœ”οƒ, is zero. πœ”οŒΏ, the trunk’s final angular velocity, is 12 radians per second (12 rad/s). The trunk takes 7.2 seconds to reach its final angular velocity from rest; we symbolize that time as 𝑑. Substituting these values into our chosen equation of angular motion, we get πœƒ=12/+0/2Γ—7.2=6/Γ—7.2=43.2.radsradssradssrad

One complete revolution of an object equals a rotation of exactly 2πœ‹ radians. Therefore, to calculate the number of revolutions made by the trunk, we divide our result in radians by 2πœ‹ radians per revolution: 43.22πœ‹/=6.875….radradrevrev

The trunk completed slightly fewer than 7 complete revolutions as it rolled. Therefore, the number of complete rotations made by the trunk is 6.

We can summarize now what we have learned about rotational kinematics.

Key Points

  • Rotations can be measured in units of radians or degrees. There are 2πœ‹ radians in one full revolution.
  • Since 2πœ‹ radians equals 360 degrees, we can convert an angle in degrees to radians by multiplying it by 2πœ‹360radiansdegrees, while we can convert an angle in radians to degrees by multiplying it by 3602πœ‹degreesradians.
  • The angle through which an object has rotated is called its angular displacement and is symbolized by πœƒ.
  • The rate at which πœƒ changes over time equals an object’s angular velocity. It is symbolized by πœ” and πœ”=Ξ”πœƒΞ”π‘‘.
  • The rate at which πœ” changes over time equals an object’s angular acceleration. It is symbolized by 𝛼 and 𝛼=Ξ”πœ”Ξ”π‘‘.
  • Angular displacement, velocity, and acceleration are used in equations of motion describing rotation under constant angular acceleration: πœ”=πœ”+𝛼×𝑑,πœ”=πœ”+12Γ—π›ΌΓ—πœƒ,πœƒ=πœ”Γ—π‘‘+12×𝛼×𝑑,πœƒ=πœ”+πœ”2×𝑑.οŒΏοƒοŒΏοŠ¨οƒοŠ¨οƒοŠ¨οŒΏοƒ

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