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Circular Motion

Centripetal acceleration and forces. Use the force's framework to solve problems easily. Banked curves...

If you’re studying AP Physics 1, you’re probably familiar with circular motion – the motion of an object moving in a circle at a constant speed. Solving circular motion problems can be challenging, but with the right approach, you can tackle these problems with confidence.

One key thing to remember is that circular motion is a type of accelerated motion. This means that the object is changing its velocity, even if its speed is constant. To solve circular motion problems, you’ll need to use F = ma – just like any other force problem.

(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:

Linear VariablesRotational Variables
∆x (displacement in meters )∆θ (angular displacement in radians)
∆v (velocity in m/s)∆ω (angular velocity in rad/s)
a (acceleration in m/s2)𝛂 (angular acceleration in rad/s2)
Table 1: Linear vs Rotational variables

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 in 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.

Rotational to LinearLinear to Rotational
∆x = ∆θ × r∆θ = ∆x/r
∆v = ∆ω × r∆ω = ∆v/r
a = 𝛂 × r𝛂 = a/r
Table 2: Linear and Rotational 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.

(5) 10 practice questions to help you master circular motion

Solve all 10 questions first. Answers are given at the end of the section.

(1) A child has a toy tied to the end of a string and whirls the toy at constant speed in a horizontal circular path of radius R. The toy completes each revolution of its motion in a time period T. What is the magnitude of the acceleration of the toy (in terms of T, R, and g)?

⇨ concepts involved: Period, Net forces, centripetal acceleration
⇨ difficulty: easy

(2) A delivery truck is traveling north. It then turns along a leftward circular curve. This the packages in the truck to slide to the RIGHT. Which of the following is true of the net force on the packages as they are sliding?

(a) There is not enough force directed north to keep the package from sliding.
(b) There is not enough force tangential to the car’s path to keep the package from sliding
(c) There is not enough force directed toward the center of the circle to keep the package from sliding
(d) The force is directed away from the center of the circle.

⇨ concepts involved: net forces, direction of centripetal acceleration
⇨ difficulty: easy

(3) A 1kg and unknown mass (M) hangs on opposite sides of the pulley suspended from the ceiling. When the masses are released, M accelerates down at 5 m/s2. What is M?

⇨ concepts involved: pulley systems, multiple mass systems
⇨ difficulty: medium

(4) Conceptual Problem: Suppose you are a passenger traveling in car along a road that bends to the left. You will feel like you are being thrown against the door. Why? What causes this force?

⇨ concepts involved: centripetal forces
⇨ difficulty: medium

(5) A 250 newton centripetal force acts on a car moving at a constant speed in a horizontal circle. If the same force is applied, but the radius is made smaller, what happens to the speed, v, and the frequency, f, of the car?

(a) v decreases and f decreases
(b) increases and f decreases
(c) v decreases and f increases
(d) v increases and f increases

⇨ concepts involved: frequency, centripetal acceleration, friction
⇨ difficulty: medium

(6) Riders in a carnival ride stand with their backs against the wall of a circular room of diameter 8.0 m. The room is spinning horizontally about an axis through its center at a rate of 45 rev/min when the floor drops so that it no longer provides any support for the riders. What is the minimum coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall?

⇨ concepts involved: linear and centripetal forces, friction
⇨ difficulty: medium-hard

(7) FRQ-Style Question. A ball of mass M is attached to a string of length L. It moves in a vertical circle and at the bottom the ball just clears the ground. The tension at the bottom of the path is 3 times the weight of the ball. All answers must be in terms of M, L, and g.

(a) What is the magnitude and direction of the net force of the ball when it is at the top?
(b) What is the velocity of the ball at the top?
(c) The string is cut exactly when the ball is at the top. How long does it take to reach the ground?
(d) How far does the ball travel horizontally before hitting the ground?

⇨ concepts involved: Newton’s laws, Kinematics,
⇨ difficulty: Hard

(8) A spacecraft somewhere in between the earth and the moon experiences 0 net force acting on it. This is because the earth and the moon pull the spacecraft in equal but opposite directions. Find the distance, D, away from Earth, such that the spacecraft experiences zero net force. NOTE: You may need the mass of the earth and moon.

⇨ concepts involved: FBD, F = ma, inclines
⇨ difficulty: medium

(9) A curve with a radius of 125 m is properly banked for a car traveling 40 m/s. What must be the coefficient of static friction for a car not to skid when traveling at 53 m/s?

⇨ concepts involved: friction, banked curves, centripetal acceleration
⇨ difficulty: medium

(10) Rotational Kinematics – 3 questions.

(a) A ball rolls down a 1.5-meter long incline from rest to 3.2 m/s. The ball has a 16.0 cm radius.
Find the angular acceleration of the ball.

(b) A CD player spins at 7329 rpm. It starts from rest and has an acceleration of 419 rad/s2.
How long does it take to reach full speed?

(c) An rotating object has a uniform angular acceleration of 5.15 rad/s2.
How long does it take to speed up from 3.33 rad/s to 33.3 rad/s?

⇨ concepts involved: rotational and linear kinematics, constant angular acceleration
⇨ difficulty: medium

Jason Kuma
Jason Kuma

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

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