# Circular Motion Speed Review

###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

###### Article Content
Circular motion is among the most confusing topics in AP Physics 1. It’s riddled with misconceptions that this article will attempt to clear up. By the end of this article you will understand how you can easily solve any circular motion problem.

### Circular Motion

The words circular and centripetal are used interchangeably.

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 thing as a centrifugal force.

#### A common misconception

Note – The terms linear velocity and tangential velocity are used interchangeably.

Acceleration 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, the velocity vector points right
• at the bottom of the motion, the velocity vector points 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.

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 + 2\alpha \Delta \theta .

#### Solving Circular Kinematic Problems

1. Identify all the given variables. You will need atleast 3 known variables and 1 unknown variable.
2. Pick your rotational kinematic formula that fits the given variables.
3. 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.

It’s important to understand that angular displacement ( \Delta \theta ) is the change in radians.

If given a displacement in revolutions, convert it to radians.

1 revolution = 2 \pi radians

For example, if an object has \Delta \theta = 30 \, \text{radians} it has made 30/2\pi 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.

Why would this be useful? Take a look (2.2) where the object traveled \Delta \theta = 30 \, \text{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, \Delta x = \Delta \theta \times r, found in Table 2 above, we are able to find the horizontal distance the bike travels (15 m).

#### Period vs Frequency

The period (T) is the time taken to complete 1 full revolution.

For example, a fan rotates 1 second per revolution.

Frequency (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}

a_c and thus F_c both point inwards (towards the center of the circle).

Keep in mind the a “centripetal force” is just a placeholder. The actual force can be any force like tension, friction, or normal force as long as it pointed towards the center of the circle.

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

1. Satellites rotating around the earth.
2. Roller coaster at the top vs bottom of a loop
3. Planes flying in a circle
4. Cars going around a flat curve
5. Cars going around a banked curve
6. Swinging on a rope and pendulums

#### 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.

1. Make an FBD. The net force should always point towards the center of the circle.
2. Use Newton’s second law to make an equation for the net force that is directed to center of the circle.
3. 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, try out Elite Physics Tutoring or other physics programs. We guarantee that we can improve your understanding and scores in less than 3 lessons!

### Practice – Try these 10 Questions

Question 1
Difficulty - Beginner
Solve Type - Mathematical
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)?
Question 2
Difficulty - Intermediate
Solve Type - Conceptual
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?
1. There is not enough force directed north to keep the package from sliding.

2. There is not enough force tangential to the car’s path to keep the package from sliding

3. There is not enough force directed toward the center of the circle to keep the package from sliding

4. The force is directed away from the center of the circle.

5. None of the above.

Question 3
Difficulty - Beginner
Solve Type - Mathematical

You are working out on a rowing machine. Each time you pull the rowing bar toward you, it moves a distance of 1.25 m in a time of 0.84 s. The readout on the display indicates that the average power you are producing is 76 W. What is the magnitude of the force that you exert on the handle?

Question 4
Difficulty - Intermediate
Solve Type - Conceptual
Suppose you are a passenger traveling in car along a road that bends to the left. Why will you feel like you are being thrown against the door. What causes this force?
Question 5
Solve Type - Proportional Analysis
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?
1. v decreases and f decreases

2. v increases and f decreases

3. v decreases and f increases

4. v increases and f increases

5. None of the above

Question 6
Solve Type - Mathematical
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?
Question 7
Solve Type - Mathematical
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. The distance between the Moon and Earth is ~3.844 x 108 m. NOTE: You may need the mass of the earth and moon. You can find this in the formula table.
Question 8
Solve Type - Mathematical
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. Give all answers in terms of M, L, and g.
Question 9
Solve Type - Mathematical
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 (µs) for a car not to skid on the same curve when traveling at 53 m/s?
###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

## Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy

Unit 5 – Momentum

Unit 6 – Torque

Unit 7 – Oscillations

Unit 8 – Fluids

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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