### Unit 3 Breakdown

You are on Lesson 2 of 4

**Unit 3.1 | Understanding circular motion and centripetal forces****Unit 3.2 | Solving circular motion problems using FBDs**(Current Lesson)**Unit 3.3 | The gravitational force****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.

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

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### 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:

- Draw an FBD
- Find [katex] F_{net} [/katex] in either direction and set it equal to [katex] ma [/katex]
- 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 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 [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)

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

### PS 2 and 3

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### 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.

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.

### 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.