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Unit 3.2 | Solving Circular Motion Problems using FBDs

car drifiting around circular track; circular motion; solving circular motion problems
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Jason Kuma

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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?
View Full Question and Explanation
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.

View Full Question and Explanation
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:
View Full Question and Explanation
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).
View Full Question and Explanation

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.

Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA

Programs

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

Reading Key

LRN
RE
PS
PQ
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White
Blue
Orange

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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)}
\omega (Angular Velocity)\text{radians per second (rad/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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