### Circular Motion

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 a things as force a *centrifugal* force.

#### A common Misconception

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

Acceleration we 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, velocity point to the right
- at the bottom of the motion, velocity points to the 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.

Linear Variables | Rotational Variables |
---|---|

∆x (displacement in meters ) | ∆θ (angular displacement in radians) |

∆v (velocity in m/s) | ∆ω (angular velocity in rad/s) |

a (acceleration in m/s^{2}) | 𝛂 (angular acceleration in rad/s^{2}) |

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 + 2a \Delta \theta .

#### Solving Circular Kinematic Problems

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

#### Using Radians

It’s important to understand that angular displacement (**∆θ**) is the* change in radians*.

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

1 revolution = 2π radians

For example, if an object has **∆θ = 30 radians** it has made 30/2π 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.

Rotational to Linear | Linear to Rotational |
---|---|

∆x = ∆θ × r | ∆θ = ∆x/r |

∆v = ∆ω × r | ∆ω = ∆v/r |

a = 𝛂 × r | 𝛂 = a/r |

Why would this be useful? Take a look (2.2) where the object traveled **∆θ = 30 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, **∆x = ∆θ × r**, found in Table 2 above, we are able to find the horiziontal distance the bike travels (15 m).

#### Period vs Frequency

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

For example, a fan rotates 1 second per revolution.

Frequency (symbol ** 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 = ma _{c}** (lowercase f is friction).

#### Common Examples of Centripetal Acceleration

- Satellites rotating around the earth.
- Roller coaster at the top vs bottom of a loop
- Planes flying in a circle
- Cars going around a flat curve
- Cars going around a banked curve
- 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.

- Make an FBD. The net force should always point towards the center of the circle.
- Use Newton’s second law to make an equation for the net force that is directed to center of the circle.
- 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, please be sure to check out my Elite Physics Tutoring program or other Physics programs. We guarantee that we can improve your understanding and scores in less than 3 lessons! Book an hour of FREE tutoring with me and try me out! No strings attached.

### Try these 10 Questions

*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*)?

There is not enough force directed north to keep the package from sliding.

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

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

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

None of the above.

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?

*v*and the frequency

*f*of the car?

*v*decreases and*f*decreases*v*increases and*f*decreases*v*decreases and*f*increases*v*increases and*f*increasesNone of the above

*minimum*coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall?

*D*away from Earth, such that the spacecraft experiences zero net force. The distance between the Moon and Earth is ~3.844 x 10

^{8}m. NOTE: You may need the mass of the earth and moon. You can find this in the formula table.

*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*.

_{s}) for a car not to skid on the same curve when traveling at 53 m/s?