This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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Preparing for the AP Physics 1 Exam

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This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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Circular Motion and Gravitation

Topics include kinematics and dynamics of circular motion, Newton’s law of universal law of gravitation, and applications of topics. You will watch 2 videos, complete 2 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

In this module we will take a look at uniform circular motion, gravitation,

and torque.

Let's start our discussion by defining what we mean by uniform circular motion.

In physics, the word uniform means that something is constant.

In this case we are analyzing objects that move

in a perfectly circular path with constant speed.

For objects moving in circular motion,

speed can be calculated using the definition from our kinematics unit.

Since speed is distance divided by time, we can divide the circumference of

the circle by the time it takes to complete that one rotation, as seen here.

The time for one rotation is called a period, and

it's represented by a capital T.

Recall that period is the inverse of frequency

which is the measure of how often a phenomena occurs per second.

In other words, frequency tells how many revolutions occur per second,

and the period tells us how many seconds it takes to complete one revolution.

This means that the units for frequency are one over seconds, also called hertz.

>> We stated earlier that an object in uniform circular motion

is traveling at a constant speed.

While the object has a constant speed,

this does not mean that the object experiences zero acceleration.

Recall that acceleration is the change in velocity over time, not speed.

Since velocity is a vector, it can change in its magnitude or direction.

Notice here that as an object moves along a circular path,

the direction of the velocity is continuously changing.

This means that an object moving in circular motion must experience

an acceleration.

But what is the direction of this acceleration vector?

Using our definition of acceleration,

we can subtract an initial velocity from a final velocity as seen here.

The resultant of this operation produces an acceleration vector

that points towards the center of the circle.

This is true for all cases of uniform circular motion.

The acceleration of the object is directed towards the center of the circle.

It is even referred to as a radial acceleration,

and is represented by the equations seen here.

In later modules we will discuss objects that travel in a circle with

changing speed.

This means that the object will also have an additional acceleration tangent

to its circular path.

In our current study of uniform circular motion,

our objects will have zero tangential acceleration.

Objects moving in circular motion are not at rest, or

moving in straight line motion at constant speed.

This means that there must be a net force present on these objects

that is directed in the same direction as the object's acceleration.

This net force is sometimes referred to as the centripetal, or center-seeking force.

This is a sum of all the forces on the object

that pull it towards the center of the circle.

As you can see, the net force and acceleration point towards the center.

The velocity is always perpendicular to these vectors, and tangent to the circle.

>> We will start our problem

solving here with an object moving in a horizontal circle.

A 0.12 kilogram ball is attached to a massless string and

spun in a horizontal circle at constant speed with radius 0.3 meters.

Calculate the acceleration of the ball if it makes four revolutions every second.

If the ball's radius of motion stays constant, but now increases

to eight revolutions per second, calculate the ball's new acceleration.

Let me first look at the variables that have been provided with this problem.

So I'm given the mass of the ball, the 0.12 kilograms.

I'm also provided with the radius of the string, which is 0.3 meters.

And now, from here, I'm asked to solve for the acceleration.

Well, let me go ahead and look at my free body diagram.

So suppose I have this object, this ball right over here.

And I know that I have gravitational force, mg,

the weight acting downwards on it.

However, this is a horizontal motion, horizontal circle.

So I'm going to apply Newton's second law to the radial direction,

which we will assume is perfectly horizontal.

Let's just suppose there's a hand here holding the string.

What I have here, my force that's acting is the tension

that's being exerted towards the hand.

So I have my tension force.

Fnet = ma, my basic law that I'm going to start with.

And, from here, I can go ahead and see if I want to solve for acceleration.

My equation for it is a = v squared over r.

Well, I don't have the velocity that this object is moving with,

but I also have another equation which is v = 2 pi r over T.

Now there's one other variable that I've been provided with that I

haven't noted yet.

This question tells me that this ball makes 4 revolutions every second.

So, 4 revolutions

per each second, well that's not the period, that is the frequency.

Frequency is how many rotations or revolutions occur per each second.

The period, however, is how many second it takes for one revolution to occur.

So this is the frequency, and there's an inverse

relationship between frequency and period,

which means that the period I have is 1 over 4, or 0.25 seconds.

I can now go ahead and plug this into my equation for velocity.

So I have v = 2 pi, r being 0.3, and

the period being 0.25 seconds.

This gives me 7.54 meters per second for

the magnitude of velocity.

I can now go ahead and use this to solve for acceleration.

If I recall, acceleration was v squared over r.

Therefore, I now have 7.54

squared over r, which is 0.3.

This provides me with an acceleration of

189.5 meters per second squared.

And since that's all the question was asking for that segment,

that's where I'll stop.

Let's look at the next portion.

If the ball's radius of the motion stays constant, but

now increases to 8 revolutions per second, calculate the ball's new acceleration.

So let's go ahead and solve for this using proportionality.

They're changing how many revolutions are occurring per second, which means that

they are changing the frequency, which means that they are changing the period.

I need to see how this impacts the speed, or the magnitude of the velocity of this

object, and then I need to relate that to how acceleration will change.

Remember acceleration is v squared over r and

v is 2 pi r over the period.

Let's consider how the period changes.

Instead of equaling 1/4

it now equals 1/ 8.

Which means that the period has been halved.

Well let's see how this would affect the velocity.

V is inversely related to the period.

Notice that I'm allowing everything else over here to equal 1 because all

these other conditions are remaining constant.

So that means that v is proportionate to 1 over 2,

which means that v changes by a factor of 2.

This makes sense, an inverse relationship.

The period is half, which means that my speed or

the magnitude of my velocity is doubled.

I can now look at how this relates to acceleration.

Looking back at the equation, again, considering that everything else

remains constant, acceleration is directly proportionate to v squared.

So if a is proportionate to v squared and v is changed by a factor of 2,

then a will be changed by a factor of 2 squared,

which means that the acceleration is quadrupled.

So my new acceleration is going to be 4 times my original acceleration.

If my original acceleration was a 189.5,

then I can go ahead and multiply that by 4,

and that gives my acceleration of

758 meters per second squared.

This could've also been solved mathematically using the equations, and

over here I'm showing you the solution that you would get if you chose to use

the equation.

So you would start with B equals 2 pi r over T, remember that

the period was halved so it went from 0.25 to 0.125.

You put that in and you get your velocity or magnitude of it.

It's 15.08 meters per second.

Then you use your equation a equals v squared over r.

Using your radius which remember stayed the same as 0.3.

You end up with the acceleration 758 meters per

second squared which is exactly what we got using proportionality.

Either way of solving the question is perfectly fine.

This just provides you with two methods.

Now the same 0.12 kilogram ball from the last example is spun in

a vertical circle of radius 0.6 meters.

Calculate the minimum speed the ball must have at the highest point so

it continues to move in circular motion.

If the ball travels at 3 times this speed at the bottom of the circle,

calculate the tension in the string.

Unlike horizontal circular motion, in a vertical circle, the speed and

direction are constantly changing due to gravity.

The ball speeds up as it falls and slows down as it rises.

Still, we can use our equation A=b^2 over r at any point and

we will specifically look at the very top of the circle and

the very bottom of the circle.

So in this question, we're asked to solve for the minimum speed that the ball must

have at its highest point so it continues to move in circular motion.

Well let's go ahead and

look at the pre body diagram of this ball at the very top of its motion.

Remember that your net force is directed towards the center of the circle.

I have mg acting on this object,

and I also have the tension force that's acting as well through the string.

So as I'm looking at this I want to solve for the critical or

minimum velocity just enough for this ball to pass through the top of

the circle while the string is still tightly drawn.

This means that tension will approach 0.

I have my equation Fnet = ma.

My net force consists of tension and mg.

Notice how these are both positive.

That's because they're following the direction of the acceleration.

Acceleration is directed towards the center, as is the net force.

And so tension and mg are both positive.

Like I said in order to solve for this minimum or

critical velocity tension will approach 0.

So what I'm left with is mg equals mg squared over r.

The mass is the same so I can go ahead and cancel that out.

And now I have g = v squared / r.

I need to solve for v, so my equation for

critical velocity reads v = root gr,

g being gravitational acceleration.

I'm using 10, you could also use 9.8.

r in this question was 0.6.

When I solve for this,

I get an answer of 2.45 meters per second.

The next part asks if the ball travels at 3 times the speed at the bottom of

the circle, calculate the tension in the string.

Okay? So let's consider this v that I just

solved for as my v1.

I need to solve for v2 now, which is the speed at the bottom of the string.

Well v2 they tell me, or

this v at the bottom, is 3 times the one that we just solved for.

So that is 3 times 2.45 meters per second which gives me 7.35 meters per second.

Now I need to solve for the tension.

Well again let's look at our free body diagram.

So at the very bottom, I have mg, but I also have tension.

Well which direction is tension going to be pointing?

The same direction as the net force is directed towards the center.

Remember this tension is exerted by the string that's spinning in

a vertical circle, so my tension is directed towards the center,

it's in the opposite direction of mg this time.

Acceleration is also directed towards the center which means tension is positive and

mg is negative.

So I write out,

Fnet = ma, T -mg = ma.

And I know that acceleration equals v squared over r so I can plug that in.

Plus mg and now I can go ahead and plug in my values The mass was .12,

and using the new velocity now, it's going to be triple about that.

Also be sure to square it.

Often times students forget that part.

Radius is .6, that remains the same.

And then mg, g being gravitational acceleration,

I'm going to use ten meters per second squared for that.

When I solved for my tension here, I get the answer of 12 Newtons.

>> Let's look at an example in which some of the forces are oriented at an angle.

A mass of 0.4 kilograms is suspended on a string that is 1.5 meters long,

and spun in a horizontal circle so

that the string makes an angle of 20 degrees with the vertical.

Calculate the tension in the string.

Next, calculate the speed of the mass in this conical pendulum.

To solve a problem like this, first I want to start with a diagram.

I have a ball that's going in circular motion in what we call a conical pendulum.

Which means that rather than being spun in a completely horizontal circle,

this ball here is being spun horizontal circle, but the string describes a cone

because this angle here, according to the problem, is 20 degrees to the vertical.

And if I want to calculate the forces and

the speed involved in this problem, I'm going to start with forces.

So, drawing the forces that act on this ball, a string can only pull.

There is a tension along the direction of the string.

And the force of gravity,

mg, always pulls toward the center of the Earth, directly downward.

And now, looking at those forces, I know those forces are balanced in the y

direction because my ball is moving in a horizontal circle.

It's not accelerating up or down.

So I could set up an expression.

Sum of forces equals the mass of that ball times the acceleration in the y direction.

And know that that acceleration will be zero.

So I can break up my forces in the y direction.

On this free body diagram now, I'm going to break it into components.

Notice my mg is already completely vertical, but my tension is not.

I can break up my tension with a horizontal component and

a vertical component.

And I noticed that this angle here,

this theta, is the 20 degrees that the string makes with the vertical.

So I'm going to use that in my equation.

In the vertical direction then I have a T Cosine of theta pulling that mass upward.

So it's a positive tension.

Downward, my downward is my negative direction, mg, so minus mg.

Those are my two forces in the vertical direction,

one positive because it pulls up, one negative because it pulls down,

should equal zero because my mass is not accelerating In the vertical direction.

So I can solve this, let's see where it takes us.

For tension, moving mg over it becomes positive.

And dividing by the cosine of theta.

Let's sub in our numbers here, the small mass 0.4 kilograms, g, we're using 10.

Cosine and the angle we said would be the angle the string makes with the vertical,

which is 20 degrees.

Make sure your calculator is in degree mode here when you use these numbers.

And I get a tension of 4.26 Newtons in that string.

And now, that's the tension pulling it diagonally.

The next question asks us is the speed that this mass needs to be traveling in

tangent to that circle as it moves in this horizontal uniform, circular motion.

So, since that's in a horizontal direction, let's see how we can set

an equation up like that, gave myself a little bit more room here but

tried to leave a little bit of my diagram.

Sum of forces in the x direction equals ma for us.

And I notice that mg in this free body diagram only pulls it vertically.

Has no horizontal component so that not going to show up.

But I do have a horizontal component of the tension.

I have this horizontal component which would be T sine of theta for

this component here, and that pulls to the center of the circle.

Remember for circular motion,

we're making the direction towards the center of the circle.

The direction of the acceleration positive.

So what I want to have is a T sine theta which is positive and

that's the only force directed to the center.

That would be what we call our [INAUDIBLE] force.

Remember [INAUDIBLE] force never shows up on our free money diagram, or an equation.

It's really a sum of forces, so

the name itself is a little misleading for some people.

That should equal M and then since this is traveling in a circle,

I know these acceleration should be V squared over R.

So, I replaced that acceleration.

So, let's look at the variables we do now, we've already solved for tension so

we know T, we already know the angle.

Again the theta is 20 degrees, the angle of the string makes with the vertical.

The mass they've already told us.

The radius here is the radius that this circle makes

as it forms its horizontal circle.

So that's something we actually solve for.

Notice though that I have a nice right triangle here.

Where this would be my radius, and this is the length of my string,

and then we have some vertical height.

So this distance r, this radius that I need, I can use trig for.

This is a right angle, I could say something like, well,

my r will equal the length of the hypotenuse times the sin theta.

In this case they told me the length of that string was 1.5 meters.

And again, our sin is 20.

So that in this problem the radius that ball makes as it goes in its horizontal

circle is 0.513 meters.

Which is what I will sub in into my expression here to find v.

So subbing in our numbers.

The tension that we found

a moment ago up here, 4.26 sin(20).

The mass of the ball now, we're just talking about forces on the ball, so

we need just the mass of the ball.

In this problem, the string is massless.

0.4.

We're looking for v, and the radius of the motion we just discovered which was 0.513.

Solving that for v, I'm going to multiply by the 0.513,

divide by the 0.4, and then take the square root.

I get a speed in the end of 1.37 meters per second.

That's the speed this ball has as it travels in the horizontal circle as

part of this conical pendulum.

The math found in banked curve problems looks very similar to conical pendulums.

A car approaches a banked curve of radius 100 meters

with a speed of 30 meters per second.

If the roadway is frictionless, perhaps it has snowed, calculate the angle at which

the road should be banked so that the car does not slide up or down the incline.

Much like the conical pendulum problem,

I'm going to start this scenario by drawing a sketch.

Where, when a car approaches a banked curve, the roadway itself is

banked upwards, and the car has a better chance of staying on the road.

So this is the bank.

The angle is what we're solving for.

What angle should this road be curved upward?

Think of a race track, so that the car will stay on,

even if there were no friction at all, between the tires and the road.

So this is the car that goes in the circle,

that means that the center of the circle that this car travels in

is somewhere over here.

Then there needs to be a net force to the left on this car as it goes in what we

could call a horizontal circle, much like the conical pendulum.

We don't want that car to travel vertically upward or downward at all.

So let's take a look at the forces on this vehicle.

It's on a roadway, it's on a surface, and we have a normal force.

Normal force, as always is perpendicular to the surface.

And we have gravity, a force of gravity pulling you straight down, MG.

When we did inclined planes we broke up our forces in a titled set of axes.

Where it made our lives easier.

We could maybe break up only one force rather than breaking up

three forces in the components.

Here we don't want to do that because we need to, in a moment,

know the net force in the horizontal direction to the center of the circle.

That's going to give us our f equals ma,

and then a equaling that v squared over r for circular motion.

We'll see that in a moment.

The trick on these conical pendulum problems and the banked curves is not to

tilt your axis in the incline, leave it horizontal and vertical.

In that case, I'm going to start setting up expressions here.

Again I know sum of forces equals MA, that's always true.

Let's do this in the Y direction, it asks us, so that the car does not slide up or

down the incline, that means the acceleration of that y direction will be

zero, it also means I need to break up my forces in the y direction.

So, the normal force is the only force not completely vertical or horizontal.

It pushes up and left on the car, and this angle theta

here will be the same as the angle here at the bottom of the incline.

So, since those two angles are the same, if I want the vertical component

of the normal force, and that would be N cosine theta,

positive because it pushes up on the car, remember up is positive, down is negative.

So minus mg which we don't have to break into its components it's entirely

vertical, and that should equal 0.

Notice that this equation looks very similar to what we got for

tension in the chronicle pendulum.

Well, that doesn't help us yet, because I don' know what the normal force is.

Even though I do know mg, that will be ten,

I don't know the angle, and I don't even know the mass of the car.

So, I have so many unknowns at this point that I can't solve.

Let's try setting up another equation that will help us get a little further.

Let's try setting up some of forces in the x direction.

So, scroll me down a little bit to give us more room.

There we go.

Sum of forces in the x direction equals ma, and

this car does accelerate in the x direction.

Remember, like we said a moment ago,

the center of the circle's over here to the left of this car.

There needs to be a net force in that direction to cause it to accelerate.

It's not at rest or moving in a straight line motion.

That means that that to the left will also be our positive direction.

We set our direction of acceleration in the positive direction.

That means N sine theta, this component here, is positive.

It points towards the center of the circle.

We don't have any other forces acting in this direction, so I can ignore those.

Mv squared over r.

We're solving for theta.

I know the speed.

They told us that.

I know the radius.

They told us that.

So why don't we take and combine this equation with this equation number.

Since I've solved equation one for normal force,

I'm going to sub that in to our equation two.

And do that right here.

That means I'm going to have an MG divided by, I'm sorry, cosine theta.

And then the sine of theta still remains.

That will equal m, v squared, over r.

Now that I've combined those two,

there are a couple of simplifications we can make.

Notice that the mass of the car shows up on both sides of the equation, so

dividing out means that they are eliminated.

I also have a trick substitution here that comes in awfully handy that the tan of

theta equals the sin of theta divided by the cosine of theta.

And I'm going to take that and sub back in here as well.

And so I end up with an expression that looks like this, g.

Tan theta equals b squared over r.

We're solving for theta, so I can rearrange this expression, and

use the inverse tangent.

Something like this.

Inverse tangent, or arc tangent, would be v squared over gr,

and that would give us theta.

I have divided by g, and then taken the inverse tangent of that number.

I can plug in the numbers that they gave us.

In this case, the velocity

that we want the car to have as it goes around the curve, 30 meters per second.

G of course for us is always going to be ten.

The radius of the curve was 100 meters.

I'm going to sub that in.

To solve this make sure again your calculator is in degree mode.

That's quite a number of degree.

And when I plugged those numbers into my calculator I got 41.99 degrees.

Of course 42 degrees would be perfect [INAUDIBLE] you could round

without a problem.

That's the angle we need that curve to be banked, so

that the car doesn't need any friction and all,

and that means it doesn't accelerate up or down on that ramp.

If the incline were steeper than that,

then the car would fall inward [INAUDIBLE] if it were less steep than that,

there's not enough net force inward on the car and it starts to slide up the ramp.

One more thing to notice here, we're used to here,

when we get to normal force on an incline plane, having normal force equal mg

cosine of theta when just a box is sliding down an incline plane.

That's not the expression we get here.

The car here actually pushes against the ground more as it goes around this curve,

that's just a simple box sliding down an incline,

you do not get the same expression.

Be careful because that is a common misconception.

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