Unit 2.B | Solving Circular Motion Problems using FBDs

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 covered in 2.A).

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 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 FBDs.

Common circular motion scenarios and their FBDs

Try drawing an FBD for the following situations. Then open the FBDs below 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)

Show the FBDs (try them first!)

\(T\)

\(mg\)

\(N\)

\(mg\)

\(N\)

\(mg\)

\(f\)

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 [katex] F_{net} [/katex] in either direction and set it equal to [katex] ma [/katex]
3. Substitute missing variables and solve

Just remember to change the type of acceleration to centripetal:

[katex] F_{net} = ma_c = m\frac{v^2}{r} [/katex]

One more formula

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

So because objects move at constant speed during uniform circular motion, we use the [katex] d = vt [/katex] formula to find the period.

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

[katex] T = \frac{2\pi r}{v} [/katex]

Note: [katex] T [/katex] (capital T) is period, whereas [katex] t [/katex] 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 [katex]T[/katex] (tension)

\(T\)

\(mg\)

\(a_c\)

\(m = 0.1\,\text{kg} \quad v = 5\,\text{m/s} \quad r = 0.5\,\text{m}\)

\(T - mg = \frac{mv^2}{r}\)

Answer: [katex] \approx 6 \, N [/katex].

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.

Lesson 2.C 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. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-3-2-solving-circular-motion-problems-using-fbds/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-3-2-solving-circular-motion-problems-using-fbds/