# Unit 3.2 | Solving Circular Motion Problems using FBDs

###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

###### Article Content
This lesson will cover how to solve all circular motion problems aka centripetal force problems.

### Unit 3 Breakdown

You are on Lesson 2 of 4

1. Unit 3.1 | Understanding circular motion and centripetal forces
2. Unit 3.2 | Solving circular motion problems using FBDs (Current Lesson)
3. Unit 3.3 | The gravitational force
4. Unit 3.4 | Combining circular motion and gravitation (satellites, orbits, and more)

#### In this lesson:

• You will draw FBDs for circular motion scenarios
• Use the FBDs to solve circular motion problems
• Cover a problem solving framework to help solve circular motion problems

### FBDs for circular motion

Just like we did for linear motion, we also draw FBDs for circular motion.

The only difference is that there should always be a net force pointing into the circular path. Why? Without this inwards net force, the object would just move along a straight path (as in covered 3.1).

More pointers for FBDs:

• Do NOT draw/label “centripetal force” as a vector. Instead draw the force causing the centripetal acceleration. For example, for a a yo-yo moving in a circle, tension force would be the centripetal force.
• Draw only forces! You do NOT need to show the tangential velocity.
• If needed, you can also lightly sketch out the “circular path” to give yourself a reference
• Don’t forget to draw weight, normal, and other linear forces in the scenario.

Still confused? Let’s practice drawing FBD.

### Common circular motion scenarios and their FBDs

Try drawing an FBD for the following situations. Then watch the video explanation to check your answers.

1. Spinning a yo-yo in a circle
2. A car traveling around a horizontal bend
3. A car going over a hill
4. Person on the bottom of a roller coaster loop
5. A spacecraft orbiting earth (bonus)

Locked Video Content

### Solving circular motion questions using a familiar framework

In the last unit we covered a problem solving framework for linear forces. You can read it here, but in short:

1. Draw an FBD
2. Find F_{net} in either direction and set it equal to ma
3. Substitute missing variables and solve

Just remember to change the type of acceleration to centripetal:

F_{net} = ma_c = m\frac{v^2}{r}

#### One more formula

Period, T , is the time is takes for an object to make one complete revolution. In the content of circular motion: Period is the time it takes to go once around the circular path.

So becuase, objects move at constant speed during uniform circular motion, we use the d = vt formula to find the period.

The distance, d is simply the circumference (aka one revolution) of the circle. The velocity v is the tangential velocity of the object. Solving for t we get:

T = \frac{2\pi r}{v}

Note: T (capital T) is period, whereas t is a general amount of time.

### Practice Question 1: A spinning yo-yo

A child spins a toy .1 kg yo-yo in a vertical circle at a speed of 5 m/s. The circular path has a radius of .5 meters. Find the tension in the string, when the toy is at the bottom of the circular path.

1st step: Draw an FBD

2nd step: Sum forces

3rd step: Re-arrange and solve for T (tension)

Answer: \approx 6 \, N .

### PS 2 and 3

Locked Video Content

### Your turn – 4 Practice Problems

If you can solve these questions, you can move on to the next section or use UBQ to find more questions.

Question 1
Difficulty - Beginner
Solve Type - Mathematical
A car travels at a constant speed around a circular track whose radius is 2.6 km. the car goes once around the track in 360 s. What is the magnitude of the centripetal acceleration of the car?
Question 2
Difficulty - Beginner
Solve Type - Mathematical

A 2.0 kg ball on the end of a 0.65 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 4.0 m/s. Find the tension in the string.

Question 3
Difficulty - Intermediate
Solve Type - Mathematical
Friction provides the force needed for a car to travel around a flat, circular race track. Answer the following:
Question 4
Difficulty - Intermediate
Solve Type - Mathematical
A 2 kg ball is swung in a vertical circle. The length of the string the ball is attached to is 0.7 m. It takes 0.4 s for the ball to travel one revolution ( assume ball travels at constant speed).

### Lesson 3.3 Preview

In the next lesson we will cover the gravitational force.

While this might seem unrelated to circular motion, the gravitational force plays a large role for objects orbiting earth — like satellites.

###### 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.

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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