Circular A

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Do curso por University of Houston System

Preparing for the AP Physics 1 Exam

36 classificações

University of Houston System

Preparing for the AP Physics 1 Exam

36 classificações

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.

Circular A31:05

Circular Motion A: New subtitles requested31:05

Circular B15:12

Conheça os instrutores

Dr. Paige K. Evans
Clinical Associate Professor
Math
Mariam Manuel
Instructional Assistant Professor
University of Houston

In this module,

we will take a look at Uniform Circular Motion, Gravitation, and Torque.

Lets 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 phenomenon.

Occurs per second.

In other words, frequency tells us 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 and

acceleration vector that points toward 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 refered to as a radial acceleration and

is represented by the equation 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 it's 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 refered 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 .12 kilogram ball is attached to a massless string and

spun in a horizontal circle at constant speed with radius .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,

the variables that I have been provided with in this problem.

So if given the mass of the ball, the .12 kilograms,

I'm also provided with the radius of the stream which is .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.

And so 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 equals 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 equals 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 equals two 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 four revolutions every second.

So four 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 seconds it takes for one revolution to occur.

So this is the frequency, and there is an inverse relationship between frequency and

period, which means that the period I have is one over four, or 0.25 seconds.

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

So I have V equals two Pi, r being .3.

And the period being .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.

Protocol, acceleration was v squared over r.

Therefore I now have 7.54 squared over r, which is .3.

This provides me with an acceleration, of 189.5 meters per second squared.

Then since that's all that 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 the new revolutions are occurring per second.

Which means 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 one fourths, it now equals one over eight,

which means that the period has been half.

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 in 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 me my acceleration of

758 meters per second squared.

This could have 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 v equals 2 pi r over T,

remember that the period was half so it went from 0.25 to 0.125.

You plug 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, it 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 three times the 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 equals v squared 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 free 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 its tension force that's acting as well through this 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 zero.

I have my equation, Fnet equals 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 energy are both positive.

[SOUND] Like I said, in order to solve for

this minimum or critical velocity,

tension will approach zero.

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

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

And now I have g equals v squared over r.

I need to solve for v, so my equation for critical velocity reads v

equals root g times r, g being gravitational acceleration.

I'm using 10. You can 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 three times this 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's 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 point?

The same direction as the net force.

It 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.

[SOUND] So I write out,

Fnet equals ma.

Positive tension, negative mg equals 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.

[NOISE] The mass was 0.12.

I'm using the new velocity now, so be careful about that.

Also be sure to square it.

Often times students forget that part.

Radius is 0.6, that remains the same.

And then ng, g being gravitational acceleration.

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

When I solve for my tension here I get 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 spin in a 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 gravity, a 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, an expression.

Sum of forces equals the mass of that ball times its 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 all ready completely vertical, but my tension is not.

I can break up my tension with a horizontal component and

a vertical component.

And I notice 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 pos, my downward is my negative direction mg and 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 0 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 mass, this 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.

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.

Giving myself a little bit more room here, but

trying 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, it has not horizontal component.

So that's 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 will 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 centripetal force.

Remember, centripetal force never shows up on a free body 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 the acceleration should

be v squared over r, so I've replaced that acceleration.

So let's look at the variables that we do know.

We 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 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 that we actually solve for.

Notice though that I have a nice right triangle here.

Where this would be my radius, 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 can say something like, well, my r will equal

the length of the hypotenuse times the sine of theta.

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

And, again, our sine of 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 sine of 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 that this ball has as it travels in

a 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 comical 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 upward so that 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 racetrack 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 a circle.

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

is somewhere over here.

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

would 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.

[SOUND] 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.

The force of gravity pulling straight down, mg.

When we did inclined planes, we broke up our forces in a tilted set of axes,

where it made our lives easier.

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

three forces into 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 axes 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 in that y direction will be 0.

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 was just up on the car.

Remember up is positive, down is negative.

So minus mg, which we don't have to break it, or 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 a conical pendulum.

[SOUND] Well, that doesn't help us yet because I don't know what

the normal force is even though I do know g, that would be 10.

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 sum of forces in the x direction.

So, scrolling down just a little bit, to give us a little more room.

There we go.

Sum of forces [SOUND] 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 will, to the left, will also be our positive direction,

reset our direction of acceleration in the positive direction.

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

It points towards the center of the circle.

We don't have any other forces acting in the x direction, and 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, [SOUND] with this equation.

Now when I, since I've solved equation one for a 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 a mg [SOUND] divided by, sorry, cosine theta.

[SOUND] And then the sine of theta still remains.

[SOUND] That will equal mv 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 trig substitution here that comes in awfully handy.

That the tan of theta [SOUND] equals the sine of theta,

divided by the cosine [SOUND] of theta.

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

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

g [SOUND] tan of theta equals v square 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, [SOUND] 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 10,

the radius of the curve was 100 meters.

I'm going to sub that in.

[SOUND] Solve this for theta, make sure again you're calculator is in degree mode,

it's asking for a number of degrees.

[SOUND] And when I plug those numbers into my calculator, I got 41.99 degrees.

Of course, 42 degrees would be perfect.

Sig figs you could round into that problem.

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

so that the car doesn't need any friction at 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, fall inward,

down the ramp.

If it were less steep than that, there's not enough net force inward on the car.

And 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 to this curve than just a simple box sliding down an incline.

You do not get the same expression.

Be careful because that's a common misconception.

[SOUND]

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