 Nerd Notes ###### Jason Kuma

Founder, Writer, Physic B.S, Business B.A USC, Fremont CA

# Circular Motion in 10 Minutes

This is everything you need to know to master circular motion in 10 minutes or less.
##### Contents

Earlier you learned that Kinematics is the study of linear motion. We also learned about how forces cause a linear acceleration. Similarly, Rotational Kinematics and Rotational Forces covers the study of circular motion and acceleration, respectively.

### (1) Two Parts to Circular Motion

There are two parts to circular motion: rotational kinematics and rotational forces.

Rotational kinematics is just like linear kinematics, but uses rotational variables as shown below

The seconds part of circular motion has to do with forces. Any object moving in the path of a circle, such as a pendulum, experiences a force called centripetal force. The other type of rotational force is called Torque (we’ll cover this in a later post).

Note that linear velocity is the same thing as tangential velocity. This basically means that the velocity at any point (of circular motion) is tangent to the circle.

#### (1.1) A common misconception

When we talk about acceleration we generally mean a change in velocity. Thus centripetal acceleration should mean that the objects velocity is changing, right?

Right and wrong. The velocity is changing direction, but not speed. Recall that velocity is a vector, thus has a direction and magnitude.

Take a yo-yo spinning clockwise with a linear velocity of 15 m/s. Since linear velocity is tangential at any given point:

• at the top of the motion, velocity point to the right
• at the bottom of the motion, velocity points to the left.

Velocity is clearly changing (in direction), thus there must be an acceleration. We call this centripetal acceleration.

### (2) Rotational Kinematics

Rotational kinematics is the same as linear kinematics, but with different variables (as listed above). See this post if you need to review linear kinematics.

To find the rotational kinematic equation, take the linear kinematic equation and replace the linear variables with rotational variables. So you would replace velocity with rotational velocity and so forth.

For example, take the linear kinematic equation vf2 = vo2 +2a∆x. The resulting rotational kinematic equation would be: ωf2 = ωo2 +2𝛂∆θ.

#### (2.1) Solving problems with rotational kinematics

First identify all the given variables. You will need at-least 3 known variables and 1 unknown variable. Then pick your rotational kinematic formula that fits the given variables. Lastly plug everything in and solve for the unknown.

This is pretty much how you would solve linear kinematics (but slightly easier). If you still need practice with linear kinematics here are 50 AP style questions to master kinematics. These advanced questions put together by college grads. If you can solve these you can solve any kinematic problem.

#### (2.2) Understanding radians

It’s important to understand that angular displacement (∆θ) is the change in radians. Thus, 1 full revolution is 2π radians. Thus if an object has ∆θ = 30 radians it has made 30/2π revolutions.

Make sure to switch your calculator from degrees to radians when solving rotational problems.

#### (2.3) Period vs Frequency

The period (symbol: T) of an object in circular motion is the time is take to complete 1 full revolution. For example, a fan rotates at .1 seconds per revolution.

Frequency (symbol: f ) is the inverse of period or f = 1/T. In other words frequency is the number of revolutions in 1 second. We call this number hertz (Hz). For example, a fan makes 10 revolutions in 1 second.

#### (2.3) Converting rotational motion to linear motion

R (radius of the circle) is the important link between rotational and linear motion.

Multiply R by the rotational displacement to get linear displacement. See the table below for all the conversions.

Why would this be useful? Take a look (2.2) where the object traveled ∆θ = 30 radians. Having distance in radians isn’t particularly useful. But what if the object in motion was a bike tire with a radius of .5 meters? Using the equation, ∆x = ∆θ × r, found in Table 2 above, we can find that the bike traveled 15 meters linearly.

### (3) Rotational Forces

In this post we will focus on just centripetal forces. We will cover torque (another rotational force) in a separate post.

A centripetal force accelerates an object rotationally. You find this force using the formula Fc = mac, where ac is centripetal acceleration ( ac = v2/r ).

Centripetal acceleration and thus Fc both point inwards (towards the center of the circle).

The key thing is to understand what causes the centripetal acceleration. For example, you spin a ball attached to a string. In this case the Tension in the string causes the centripetal acceleration.

#### (3.1) Solving Centripetal force problems

Solve these problems like any regular force problem. If you need a refresher on solving forces linearly, skim through Forces in 10 minutes or less.

1. Make an FBD
2. Find the net force that causes the centripetal force
3. Solve the equation

For example, take a car driving around a flat curve. The centripetal forces comes from only from the friction of the tires. Thus to find the frictional force we use the equation f = mac. (lowercase f is friction)

#### (3.2) Common examples of centripetal acceleration

1. Satellites rotating around the earth.
2. Roller coaster at the top vs bottom of a loop
3. Planes flying in a circle
4. Race cars going around a flat curve vs banked curve.

Note that on banked curves (where the road is tilted at an angle), you don’t need always need friction for centripetal acceleration. Instead, centripetal acceleration, results from a component of the weight of the car.

For example, think of the car loosing friction on a banked curved. The car will continue to travel around the curve, because the weight of the car pulls it inwards.

### (4) Extra Help!

That wraps up circular motion! If you still need help with on the topic, please be sure to check out my Elite Physics Tutoring program or other Physics programs. We guarantee that we can improve your understanding and scores in less than 3 lessons! Book an hour of FREE tutoring with me and try me out! No strings attached.

## More Related Content

### Torque in 10 Minutes

This is a speed review of Torque for AP Physics. Test yourself with 60 AP-style questions designed to challenge your understanding.

### Forces in 10 Minutes

Review forces in 10 minutes or less + 60 questions to help you master forces.

### Kinematics in 10 mintues

Review kinematics in 10 minutes or less + 60 questions to help you master kinematics. 