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### Uniform Circular Motion

• The distance around a circular path is the circumference.
• Frequency is the number of revolutions or cycles which occur each second. The symbol is f, and the units are 1/s, or Hertz (Hz).
• Period, the inverse of frequency, is the amount of time it takes for each revolution or cycle. The symbol for period is T, and units are seconds (s).
• Objects moving in uniform circular motion (circular motion at constant speed) are accelerating because their direction is always changing, causing their velocity to continuously change.
• The direction of this acceleration is toward the center of the circular path. This type of acceleration is known as a centripetal acceleration (center-seeking). The force causing a centripetal acceleration is known as a centripetal force.

### Uniform Circular Motion

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Objectives 0:08
• Uniform Circular Motion 0:42
• Distance Around the Circle for Objects Traveling in a Circular Path at Constant Speed
• Average Speed for Objects Traveling in a Circular Path at Constant Speed
• Frequency 1:42
• Definition of Frequency
• Symbol of Frequency
• Units of Frequency
• Period 2:04
• Period
• Frequency and Period 2:19
• Frequency and Period
• Example I: Race Car 2:32
• Example II: Toy Train 3:22
• Example III: Round-A-Bout 4:07
• Example III: Part A - Period of the Motion
• Example III: Part B- Frequency of the Motion
• Example III: Part C- Speed at Which Alan Revolves
• Uniform Circular Motion 5:28
• Is an Object Undergoing Uniform Circular Motion Accelerating?
• Direction of Centripetal Acceleration 6:21
• Direction of Centripetal Acceleration
• Magnitude of Centripetal Acceleration 8:23
• Magnitude of Centripetal Acceleration
• Example IV: Car on a Track 8:39
• Centripetal Force 10:14
• Centripetal Force
• Calculating Centripetal Force 11:47
• Calculating Centripetal Force
• Example V: Acceleration 12:41
• Example VI: Direction of Centripetal Acceleration 13:44
• Example VII: Loss of Centripetal Force 14:03
• Example VIII: Bucket in Horizontal Circle 14:44
• Example IX: Bucket in Vertical Circle 15:24
• Example X: Demon Drop 17:38
• Example X: Question 1
• Example X: Question 2
• Example X: Question 3
• Example X: Question 4

### Transcription: Uniform Circular Motion

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about uniform circular motion.0004

Our objectives include calculating the speed of an object traveling in a circular path or portion of the circular path.0008

Calculating the period and frequency for objects moving in circles at constant speed.0015

Explaining the acceleration of an object moving in a circle at constant speed.0020

Solving problems involving calculations of centripetal acceleration.0024

Defining centripetal force and recognizing that is not a special kind of force0030

but that is provided by forces such as tension, gravity, and friction.0034

Solving problems involving calculations of centripetal force.0038

Uniform circular motion, objects travel a circular path at constant speed, when they do that we call that uniform circular motion.0043

The distance around the circle is its circumference.0052

If there is a circle and we call the distance from the center to the edge of the radius, then circumference is 2π r,0055

the distance around the outside.0063

If instead, you define a diameter through the center point across a circle as your diameter then circumference is just π × D.0066

The average speed formula that we talked about from our kinematic section still applies here.0077

Average speed is distance traveled / time but for something moving in a circle once around the circle,0082

it would be 2 π r the circumference ÷ the time it takes to go once around the circle.0087

If it is once around the circle, you can say that is 2 π r ÷ period, where the period is the time for one complete cycle or revolution.0092

Frequency is the number of revolutions or cycles which occur each second.0103

It gets the symbol f and the units are 1/s, which are also known as a Hertz abbreviated capital Hz.0107

Frequency is the number of cycles per seconds or the number of revolutions per second.0117

Corresponding to frequency, we have the period that is the time it takes for one complete revolution or cycle.0123

Its symbol is T and the units are seconds.0129

T period time for one cycle is the time for one revolution.0133

Frequency in period are closely related.0138

The frequency is 1/ the period and the period is 1/ the frequency.0142

1 When you know one, you can easily find the other.0146

A couple of examples here starting off by talking about a race car.0151

The combined mass of a race car and its driver is 600 kg.0155

Traveling at constant speed the car completes one lap around a circular track of radius 160m in 36 s, calculate the speed of the car.0159

A radius of 160 m, average velocity is distance travel ÷ time.0170

Its distance traveled is once around so that is a circumference ÷ time or 2 π × the radius 160 m/36 s which is 27.9 m/s .0177

A very simple example but probably worth getting a good solid foundation in the basics before we move further.0195

Taking a look at a toy train, a 500g toy train completes 10 laps of its circular track in 00:01:40.0202

If the diameter of the track is 1m, find the trains period T and its frequency f.0210

Period is going to be, it takes it 100s or 1min and 40s to do 10 laps.0219

Period being the time for 1 lap or 1 revolution is just going to be 10 s.0227

Once we know that, the frequency becomes 1/ the period or 1/10s which could be 0.1 Hz.0234

Let us take a look at a more detailed example but still pretty straightforward.0246

Allen makes 38 complete revolutions on the playground around about 30s.0250

If the radius of the roundabout is 1 m, determine the period of the motion.0256

The frequency of the motion, the speed at which Allen revolves, and how sick he is when he is out.0260

A, period of the motion, the period it takes him 30s to do 38 revolutions, the period is 0.789s.0267

The frequency is just 1/ the period which would be 1/0.789s or 1.27 Hz and the speed at which Allen revolves.0284

Average speed is distance travel ÷ time.0303

The distance he travels is 2 π × the radius 1m, he does that 38 times.0306

To do that 38 times, it takes him 30s so the speed would be 7.96 m/s.0315

A little more on uniform circular motion.0330

Is an object undergoing uniform circular motion accelerating?0332

We draw our circle for something moving at constant speed and is it accelerating?0337

It is kind of a trick question.0346

If we draw the velocity along the path, notice that it is changing direction depending on where it is at.0348

And because velocity is a vector, it has direction and acceleration is change in velocity.0355

It has a vector, because you have a change in direction, you have a change in velocity.0364

Therefore, yes you are accelerating.0368

Yes, it is accelerating due to that change in direction even though the speed is staying the same.0371

How do you determine the direction of that centripetal acceleration?0380

Here, we have a look at an object moving counterclockwise around the circle.0386

At this point, it is instantaneous velocity is up and a little bit later its velocity is in that direction.0391

If we want to find its acceleration, that will be the change in velocity ÷ time which will be final velocity - initial velocity ÷ time.0398

Velocity final - velocity initial that is the same as final velocity + negative initial velocity.0412

Let us see if we can draw it to see the direction of the change in velocity.0422

Our final velocity, this vector here in green I’m going to try and reproduce here.0427

It is sliding over but something like that.0432

If our initial velocity is up, negative initial velocity must be down from that point.0435

Something like that, where this is VF this is – VI.0447

To add those vectors, we draw a line from the starting point of the first to the ending point of our last.0452

That is going to be the direction of our acceleration vector.0460

If we look here by these, if we draw that same vector here, we are pointing toward the center of the circle.0464

That is why I refer to it as a centripetal acceleration.0470

Centripetal acceleration, oftentimes abbreviated ac because centripetal means center seeking.0474

It always points towards the center of the circle, even though it is moving in a constant speed the object going0490

in uniform circular motion is constantly accelerating towards the center of the circle.0496

How do you find the magnitude of acceleration?0503

The magnitude is straightforward.0506

That is the square of the speed ÷ the radius of the circle, AC =V² /r.0508

You got to know that one.0515

Let us do another example here.0518

Miranda drives a car clockwise around a circular track of radius 30m.0521

It is a circle that you get the idea 30m.0528

She completes 10 laps around the track in 2 minutes.0530

Find Miranda’s total distance traveled in the average speed and centripetal acceleration.0534

Distance traveled is going to be the distance around the circle × 10 laps or 2 π × the radius 30m × 10 laps which will be about 1885m.0540

To find her average speed, average speed is distance over time.0561

That will be 1885m and it took her 2 minutes or 120s to do that, which is 15.7 m/s.0567

Centripetal acceleration, that is going to be the square of speed ÷ the radius which will be 15.7m/s² ÷ 30m or 8.22 m/s².0581

Let us take a look now at what causes the centripetal acceleration.0607

We have acceleration, we must have a force.0611

We are going to call that a centripetal force.0614

If an object is traveling in a circle, it is accelerating towards the center of the circle, we have to find that.0617

For an object to accelerate there must be a net force.0622

This is what we call a centripetal force, it is a center seeking force.0625

Centripetal points towards the center in order to cause a center seeking acceleration.0630

When we talked about net force in the x direction, caused an acceleration0637

in the x direction or net force in the y direction cause an acceleration in the y direction,0642

we can also note now the net force towards the center of a circle causes an acceleration towards the center of the circle.0650

A part to note here, as centripetal force is not a new force.0660

It is just a label placed on an existing force when it is directed to the center of circle.0665

Tensions can cause a centripetal force, gravity, even friction, can provide us centripetal force.0670

A centripetal force is not any magical new force that just comes into existence because something was in a circle.0676

Something has a force towards the center of the circle that causes it to move in a circle.0682

We will call that forces centripetal force.0687

It is just a more generic label for any force going towards the center of the circle.0689

As such, you do not want to label anything FC or F centripetal on a free body diagram.0693

Be more specific, write down what it is that is causing that force.0700

Calculating centripetal force.0706

If net force is mass × acceleration, net force for the center of the circle is mass × acceleration towards the center of circle.0709

We know that centripetal acceleration is V² /r.0720

We can write net force must be equal to MV² /r, that easy.0726

If you want a look at units, the units of force are going to be kilogram m² / s² mass × speed² ÷ distance which is going to be kilogram m/s²,0737

which is our definition of a Newton, the unit of force.0754

Let us take another example.0761

If the car is accelerating, is its speed increasing?0762

That depends.0767

If you think about that, we could have a car traveling with some velocity to the right and it could be accelerating to the right.0767

If that is the case, yes its speed is going to increase.0780

On the other hand, we can have a car with some velocity to the right and acceleration to the left.0785

Is its speed increasing?0797

No, it is decreasing.0800

Or we could take a look at a car traveling in a circular path with some centripetal acceleration0802

that is accelerating but it is moving at constant speed.0811

If the car is accelerating is its speed increasing?0815

Perhaps, it could be but not necessarily.0819

In the diagram below, the car travels clockwise a constant speed in a horizontal circle.0825

At the position shown in the diagram which air indicates the direction of the centripetal acceleration of the cart.0830

It is got to be A toward the center of the circle.0837

Here we have a ball attached to a spring to a string move at constant speed in a horizontal circular path.0844

The target is located near the path of the ball is shown.0850

At which point along the balls path should the string be released removing the centripetal force if the ball is to hit the target?0854

If you want the ball to hit the target, he would release it at the point when it is instantaneous velocity0863

is tension to the circle which would be right at B.0868

The moment you get rid of that centripetal force, it no longer has a centripetal acceleration.0873

Therefore, it travels in the straight line by Newton’s first law.0878

Let us take a look at the bucket swung in a horizontal circle.0885

The diagram shows a 5 kg bucket of water swung a horizontal circle of radius 0.7m at the constant speed of 2 m/s.0888

What is the magnitude of the centripetal force on the bucket of water?0896

The centripetal force, the magnitude of the centripetal force will be MV² /r or 5 kg × its speed 2m /s² ÷ 0.7m which is 28.6 N.0900

What happens if we do this in a vertical circle?0924

The diagram now shows the same bucket, same radius, constant speed 3 m/s,0927

find the magnitude of tension of a string at the top of the circle and at the bottom of the circle.0931

Let us draw our free body diagram for the top and we will start with that as our analysis.0937

We have a tension pulling down, strings can only pull, and we have the weight MG.0943

When we write Newton’s second law, I am going to write Newton’s second law in this centripetal direction for the center of the circle.0950

Towards the center the circle is down so we have T + MG and that must equal mass × acceleration or MV² /r.0957

Which implies then the tension must be MV² /r - MG which will be 5 kg × 3 m/s² ÷ the radius 0.7m - 5 × 10 or T =14.3 N.0968

Let us do the same analysis down here when the bucket is at the bottom of the circle.0992

At the bottom, our free body diagram looks a little different.1000

Our tension is pointing up, gravity still pulls down, and we will write Newton’s second law equation1005

f net C = T is pointing towards the center of the circle so that is positive.1011

Gravity MG is pointing away from the center of a circle so that is negative.1017

T - MG =MV² /r.1021

When we solve for tension, T =MV² /r + MG which is going to be 5 × 3² / 0.7 + 5 × 10.1026

Therefore, the tension now in the string is 114.3 N.1043

Quite a difference.1053

Let us take a look at the demon drop problem.1058

Put up this after an amusement park ride at an amusement park that used to visit when I was a kid.1061

The diagram shows at the top of you the 65 kg student here at point A on an amusement park ride.1069

The ride spins a student in a horizontal circle of radius 2 1/2 m at the constant speed of 8.6 m /s.1076

While that is happening, the floor is lowered and the student remains against the wall without falling to the floor.1083

There is no floor there but the student remains against the wall.1089

First off, draw the direction of the centripetal acceleration of the student on the diagram.1092

Centripetal acceleration, center seeking, nice easy towards the center of the circle.1097

There we go for that part.1102

For part 2, determine the centripetal acceleration of the student and the centripetal force acting on the student.1106

Centripetal acceleration is V² /r so that is just going to be 8.6 m/s² ÷ our radius is 2.5m which will be 29.6 m/s².1113

If we want the centripetal force, the centripetal force is going to be MAC which is 65kg × centripetal acceleration 29.6 m/s² or 1924 N.1133

Some force is pushing the student or pulling the student toward the center of the circle with 1924 N of force.1153

Number 3, what force keeps the student from sliding to the floor?1163

Let us draw a picture.1169

Here is our student against the wall, is it spinning in the direction kind of like that.1170

If I draw a free body diagram from the side, we have weight down, normal force from the wall, and we must have force of friction.1177

What keeps the student from sliding to the floor?1190

It must be the force of friction.1193

How do you get a formula for force of friction?1195

Remember, force of friction is μ × the normal force.1197

You must have a big normal force and that is what is causing the centripetal force is the normal force.1200

Force of friction keeps a student from sliding.1209

What is the minimum coefficient of friction between the student and the wall required to keep the student from sliding down the wall?1214

In this case, in order to not slide down the wall, force of friction and the weight of the student have to be balanced.1221

Force of friction must equal the weight of the student.1227

We have just said force of friction is μ × the normal force.1231

μ × the normal force must equal MG which implies then that μ must equal MG / the normal force.1237

We also know for all of this to work that the normal force is providing the centripetal force which must be equal to mac.1247

We can write that μ =MG / mac which is just G /centripetal acceleration or 10/ centripetal acceleration was 29.6.1259

We find that we have a μ of 0.34, that is the coefficient of friction that we need in order to keep the student from sliding down the wall.1275

Hopefully, that is a pretty good refresher of fundamentals on uniform circular motion.1287

Thanks so much for joining us at www.educator.com and make it a great day everyone.1292