Circular Motion
There are two parts to circular motion: circular kinematics and circular forces.
Circular kinematics utilizes the same big 4 linear kinematic equations, but with different variable (as you’ll see below).
Circular forces also called centripetal forces deals with any object moving in the path of a circle, such as a pendulum, satellite, or rollercoaster.
There is no such a things as force a centrifugal force.
A common Misconception
Note – The terms linear velocity and tangential velocity are used interchangeably.
Acceleration we generally refers to a change in speed. When it comes to uniform circular motion, the velocity is changing in direction, but not in speed. Hence the term “uniform”.
For example, imagine a yo-yo spinning clockwise with a constant linear velocity of 15 m/s.
- 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 resulting in a centripetal acceleration. Without a centripetal acceleration, an object would just continue moving in a straight line.
What happens if the speed also changes? We call this non-uniform circular motion and typically cover this in AP Physics C.
Circular Kinematics
Circular kinematics is the same as linear kinematics, but with rotational variables listed below.
| Linear Variables | Rotational 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) |
To find the circular kinematic equation, take the linear kinematic equation and replace the linear variables with rotational variables.
For example, take the linear kinematic equation (v_f)^2 =(v_0)^2 + 2a \Delta x . Swap out the linear variables for the corresponding rotational variables. The resulting rotational kinematic equation would be: (\omega_f)^2 =(\omega_0)^2 + 2a \Delta \theta .
Solving Circular Kinematic Problems
- Identify all the given variables. You will need at-least 3 known variables and 1 unknown variable.
- 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.
Using Radians
It’s important to understand that angular displacement (∆θ) is the change in radians.
If given a displacement in revoution, convert it to radians.
1 revolution = 2π radians
For example, if an object has ∆θ = 30 radians it has made 30/2π revolutions and vice-versa.
Converting from Rotational to Linear Motion
The value of R (radius of the circle) is the bridge between rotational and linear motion. See the table below for all the conversions.
| Rotational to Linear | Linear to Rotational |
|---|---|
| ∆x = ∆θ × r | ∆θ = ∆x/r |
| ∆v = ∆ω × r | ∆ω = ∆v/r |
| a = 𝛂 × r | 𝛂 = a/r |
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 are able to find the horiziontal distance the bike travels (15 m).
Period vs Frequency
The period (symbol T) is the time taken to complete 1 full revolution.
For example, a fan rotates 1 second per revolution.
Frequency (symbol f ) is the inverse of period or
f = \frac{1}/{T}
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 or 10 Hz.
Circular Forces
A centripetal force accelerates an object rotationally.
F_c = ma_c , where a_c is centripetal acceleration a_c = \frac{v^2}{r} .
Centripetal acceleration and thus Fc both point inwards (towards the center of the circle).
Keep in mind the a “centripetal force” is just a general force. The actual force that causes the circular motion can be any force like tension, friction, or normal force.
For example, imagine 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).
Common Examples of Centripetal Acceleration
- Satellites rotating around the earth.
- Roller coaster at the top vs bottom of a loop
- Planes flying in a circle
- Cars going around a flat curve
- Cars going around a banked curve.
Solving Centripetal Force Problems
Solve these problems like any regular force problem. If you need a refresher on solving linear forces, skim through Forces Speed Review.
- Make an FBD. The net force should always point towards the center of the circle.
- Use newton’s seconds law to make an equation of the net force directed to center of the circle.
- Replace any unknowns and solve the equation
Below are 10 questions to help you apply circular motion. For even more questions check out the Centripetal Motion UBQ.
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.
10 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
