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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From the course by University of Houston System

Preparing for the AP Physics 1 Exam

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University of Houston System

35 ratings

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.

From the lesson

Rotational Mechanics

Topics include torque, rotational kinematics and energy, rotational dynamics, and conservation of angular momentum. You will watch 3 videos, complete 3 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 video,

Â we will continue to compare translational motion with rotational motion.

Â We have already seen that an object moving with a linear speed will have

Â a translational kinetic energy where K equals one-half mv squared.

Â There is also energy associated with the rotation of an object.

Â The energy of a rotating body can be written as K equals one-half,

Â I omega squared.

Â Notice the similarities again between the two equations.

Â This energy is also measured in joules and is still a scalar value.

Â Let's look at an example in which an object has rotational and

Â translational kinetic energy.

Â Suppose a 3 kilogram ball, of radius 0.05 meters, is rolling

Â with a linear speed of 2 meters per second without slipping on a horizontal surface.

Â Calculate the translational and rotational energy of the ball as it rolls.

Â Suppose this object, before reaching the horizontal surface,

Â was released from rest at the top of an inclined plane.

Â From what vertical height was the ball released so

Â that it reaches this final speed at the bottom of the incline?

Â Okay.

Â In this problem,

Â we have a ball rolling along a horizontal surface without slipping, and

Â it tells us that it has a linear speed of 2 meters per second as it does so.

Â So notice that this object, this ball, is not only rolling, but

Â also physically moving forward.

Â That means it has two types of energy.

Â It's going to have a linear kinetic energy and a rotational kinetic energy.

Â So that translational kinetic energy,

Â that linear type that we've seen before looks something like one-half mv squared.

Â And for us on this problem, that's relatively straightforward.

Â They gave us those values.

Â One-half the mass of this object, 3 kilograms, and the linear speed,

Â 2 meters per second, you need to make sure it's a square of that term.

Â So the translational kinetic energy,

Â the fact that this object is actually translating, moving from one location to

Â another, looks to me to be about 6 joules, that's a scalar.

Â It is positive energy, you can't have negative kinetic energy.

Â But since this object is rotating, it's rolling along the ground,

Â there's also some energy in that rotation.

Â So to solve for that then, our equation looks very similar.

Â And so there's a one-half, there's a term that represents inertia, like mass,

Â except it's our rotational moment of inertia,

Â which we'll have to know what that equation is for, specifically, a sphere.

Â In this problem, it's a ball.

Â And then our omega squared, our velocity squared,

Â except this is our angular velocity in radians per second.

Â So I can go ahead and start to analyze this a little bit.

Â My moment of inertia, watch your fractions here, because it's a ball.

Â The moment of inertia of a ball is two-fifths mr squared of that ball.

Â And then we come to this omega squared term.

Â Well, remembering your bridge equations, that will come in handy here,

Â because the ball is not slipping.

Â So I know that there is a relationship between how fast it spins and

Â how fast it moves forward.

Â I also was never told what the angular velocity was, so

Â this will be a great way to get to that.

Â I'm going to go ahead and sub that in, v over r, to replace my omega.

Â And I need to make sure that that's squared.

Â So it looks to me that our twos will cancel out, there's one in

Â the numerator and one in the denominator, and also looks like our rs will cancel out

Â because I have two of those in the numerator and two in the denominator.

Â And so now I can sub in my numbers for this.

Â Notice it looks a lot like now, our translational kinetic energy,

Â except it's one-fifth mv squared, which is what's left over.

Â So our rotational kinetic energy will be one-fifth the mass of this object,

Â 3, the linear velocity, 2, which is still squared,

Â gives me a final rotational kinetic energy of about 2.4 joules.

Â Notice it has more linear kinetic energy than rotational kinetic energy.

Â Now the problem's not quite done at this point.

Â It then asks,

Â suppose that this ball that's rolling started at rest at the top of an incline.

Â It wants to know how high we'd have to release this ball, so

Â that it reaches this speed that we have set up.

Â So I'm going to draw a quick diagram.

Â Here's the ball at the top.

Â I'm looking for this vertical height, h, so that by the time it reaches this

Â point down here, it's moving at 2 meters per second, like we said earlier.

Â That looks to me like a conservation of energy problem.

Â We know that it's going to have potential, gravitational potential energy at

Â the beginning, and then it's going to have two forms of kinetic

Â energy at the bottom, a rotational and a translational.

Â So my fundamental principle here to solve a problem like that is that the energy,

Â all of the energy added up before,

Â will equal all of the energy added up at the end of the problem.

Â So like we said, there's a gravitational potential energy at the beginning,

Â a translational final at the end.

Â We're setting our 0 for gravitational potential energy down here at the ground,

Â so there's no more gravitational potential energy, just these two.

Â In fact, on the right hand side, I already know these numbers.

Â We just calculated those.

Â That's our 6 joules and our 2.4 joules.

Â So I already know that the right hand side of this equation will be 8.4 joules.

Â What I'm solving for now is the height.

Â So that would be our mgh, which we already know the mass of the object, as well.

Â And g, for us, in this scenario is 10, so 3, 10,

Â solving for h, the height from which we would have to release this ball,

Â so that it would reach that speed, not very tall,

Â 0.28 meters, about 28 centimeters.

Â >> This question, I am provided with the mass, which is 3 kilograms,

Â the radius of this wheel, which is 0.4 meters.

Â It's a thin hoop so I know that if I need to use moment of inertia, I'm

Â going to use the equation I equals MR squared.

Â So that part's really important for us, as well.

Â And then next they tell me that the cyclist is moving forward with

Â a linear speed of 3 meters per second.

Â So that's v.

Â She applies a constant torque of 0.5 Newton meters.

Â And then finally, they tell us the linear distance that's traveled, she travels

Â a distance of 50 meters, and they want us to solve for the final angular velocity.

Â So I don't have time, and remember our equation kinematics,

Â where we didn't have time and we didn't need time, we didn't have to solve for

Â it, so we used our equation vf squared equals vf squared plus 2ax.

Â Well, I'm going to use my angular form of that equation over here.

Â So I've got final angular velocity squared equals initial angular velocity

Â squared plus 2 angular acceleration multiplied by angular displacement.

Â Now there are several variables in here that I don't know before I can go ahead

Â and start solving for my actual unknown.

Â So let's go ahead and first solve for our initial angular velocity.

Â So in order to solve for

Â this I'm going to use my initial linear velocity divided by radius.

Â So this is the 3 meters per second divided by 0.4,

Â and I end up with 7.5 radians per second.

Â The next thing that I'm going to go ahead and solve for is the angular displacement.

Â So, the way that I'm going to go ahead and do that is I will use the equation,

Â angular displacement equals x over r.

Â So where x did, x denotes the distance that's traveled around the circle and

Â r represents the distance from the circle.

Â [NOISE] So if you wanted to see it something like this over here

Â I'm trying to solve for theta.

Â I've got r and I have the distance traveled around the circle, as well.

Â So using that I can go ahead and solve for theta.

Â It gives me 50 over 0.4.

Â This gives me 125 radians for my angular displacement.

Â Now the next thing that I need is the angular acceleration.

Â So this is where I will use my equation [SOUND] sum of

Â my torque equals moment of inertia multiplied by angular acceleration.

Â I know that the total torque on the system was the 0.5 Newton meters.

Â As we mentioned earlier, this is a thin hoop, so for

Â moment of inertia we're going to use m times r squared.

Â The mass was 3 kilogram, radius is the 0.4.

Â From here I can go ahead and solve for the angular acceleration, and

Â this gives me 1.04 radian per second squared.

Â Now that I have this part, I can go ahead and

Â travel back to my initial equation that I knew I was going to use in order to

Â solve for what this question was looking for which is the final angular velocity.

Â [SOUND].

Â Oops.

Â I accidentally put x there.

Â It's a habit.

Â It's theta, if we're looking at our angular variables here.

Â So now, I can go ahead and

Â plug in [SOUND] 7.5 squared [SOUND] plus 2,

Â 1.04 for my angular acceleration,

Â 125 for the angular displacement.

Â Be very careful at the end when you solve for your final that you make

Â sure to take the square root because this is omega squared out here.

Â I get an answer of 17.78 radians per second.

Â And, that's my final answer.

Â Angular momentum, like linear momentum, is a conserved

Â quantity where no significant outside torques act upon a system.

Â Angular momentum like linear momentum is a conserved quantity where no significant

Â outside torques act upon a system.

Â Typically we apply conservation of angular momentum to scenarios involving

Â collisions, explosions, or orbital motion.

Â We can calculate the angular momentum of an object by using the equation,

Â angular momentum equals moment of inertia multiplied by angular velocity.

Â Angular momentum is a vector and is measured in units,

Â kilograms meters squared per second, or Newton meters seconds.

Â As before, we will see the counterclockwise direction as positive.

Â Here's an example.

Â All right, so in this question you have a planet that is moving in

Â a circular orbit around a star.

Â So that's important because it's not moving around its own axis,

Â it's moving around the star.

Â So when I need to solve for the moment of inertia for this,

Â I'm going to use my equation for a thin hoop which is MR squared.

Â All right, let's see what else they give us.

Â They tell us that this planet is on average 1 times 10 to the 11th

Â meters from the star.

Â Well that's the dis, distance over here from between the planet and the star.

Â That's my R, my radius.

Â [SOUND] So I'm going to make a note of that as well.

Â Then they tell me that it takes 300 Earth days to complete one revolution.

Â Well the time and seconds that it takes complete one revolution,

Â that would be the period.

Â So we're going to see how we'll use that in just a moment here.

Â They want us to solve for the angular momentum of

Â the planet around the star if the mass of the star

Â is 4 times 10 to the 24th kilogram.

Â All right, so let's go ahead and look at what we have over here.

Â Well the angular displacement represents the distance that's

Â traveled around the circumference.

Â That is 2pi r over the radius.

Â Well, when this becomes simplified,

Â we know that angular displacement equals 2pi for one revolution.

Â So, I can go ahead and use that in order to solve for my angular velocity.

Â My angular velocity, in this case, I'm going to use 2pi over the time

Â that it takes in seconds to complete this one revolution.

Â So think back to this as your basic in

Â kinematics your velocity being your distance over time.

Â Your speed being your distance over time.

Â So now I can go ahead and plug in values, 2pi over,

Â remember, I need to use this time in seconds, so

Â I'm going to have to convert the 300 days.

Â There are 24 hours in one day,

Â [SOUND] and 3,600 seconds in one hour.

Â So, when I go ahead and solve for all of this,

Â I end up with my angular velocity of 2.42

Â times 10 to negative 7 radians per second.

Â Now this is important because I'm trying to solve for angular momentum,

Â and that is, angular momentum equals your moment of inertia multiplied by omega.

Â And earlier, we talked about what we would use for moment of inertia.

Â We're going to use our thin hoop equation, the MR squared.

Â So, and then now we can go ahead and

Â plug in values over here [SOUND].

Â The mass, remember they gave you in the question, that's 4 times 10 to the 24th.

Â [INAUDIBLE] clear.

Â [SOUND] Multiply

Â this by omega,

Â which is 2.42

Â times 10 to

Â the negative 7.

Â And when you go ahead and plug this into your calculator, you

Â should get a value of your annual momentum to be 9.68 times 10 to the 39th power.

Â And the units I'm using here are Newtons, times meters, times second.

Â [SOUND] And so that's my final answer there.

Â Now, that's part A.

Â The next part wants me to solve for the frequency.

Â Well, if I've figured out the amount of seconds that it took for

Â that one revolution.

Â Then I know that frequency is the inverse of the period,

Â 1 over T, so I can go ahead and solve for it.

Â And again, I'll just show you the steps again.

Â It would be the 1 revolution over 300 days.

Â 24 hours in one day and

Â 3,600 seconds in one hour.

Â And when you solve for this,

Â you can go ahead and get your frequency,

Â which is 3.86 times 10 to the negative 8th power.

Â Remember frequency is in hertz.

Â And that's our final

Â answer for part B.

Â >> It's important to remember that momentum is conserved

Â when there are no outside forces affecting a system.

Â And so, when we're talking about an asteroid orbiting a sun or a star,

Â in this case.

Â Well this will be a more elliptical orbit than, let's say, a planet,

Â going around a star.

Â So let's say that this is our asteroid, this can do something like this.

Â We know from Kepler's laws, that we have here.

Â You're farther away, it should go slower.

Â When it's over here it's closer, it should be going faster.

Â The area swept out by this asteroid should be equal in periods of time.

Â That's what one of our Kepler's laws told us.

Â Well, angular momentum is conserved in this scenario because there aren't any

Â measurable or significant forces acting from the outside.

Â The only real forces that we have to worry about are the ones in between

Â the asteroid and the sun itself, those equal and opposite forces.

Â So, angular momentum is a great way to solve a problem like this.

Â This is similar to the issue of when you have the skater

Â bringing her arms in in a rotation.

Â She will speed up in her spin and her angular velocity,

Â because of the conservation of the angular momentum.

Â And so, solving a problem like this,

Â I'm going to start off with our fundamental principle.

Â Our angular momentum before, all of those angular momentums added up,

Â will equal a final angular momentum in the whole system.

Â Now, we're talking about this asteroid orbiting the sun.

Â So the only thing that's really moving in our reference frame here is going to be

Â that asteroid.

Â So we have a certain moment of inertia and angular speed at the beginning.

Â And it has a a certain moment of inertia and angular speed at the end.

Â Well we're talking about this asteroid orbiting this point outside of itself,

Â it's relatively speaking in comparison.

Â It's a small point object as it travels this great distance around the sun.

Â So I'm going to use our moment of inertia for point mass, which is m r squared,

Â which is a great approximation in this scenario.

Â So, the mass of the asteroid and it's at sum first or initial radius squared.

Â And I want to do this in terms of linear speed because it's giving me linear

Â speeds.

Â So I also want to use my bridge equation, v equals r omega.

Â So I'm going to sub that in as well.

Â We've got a v1 divided by r1.

Â On the other side, it's going to look very similar.

Â There's now the mass, which hasn't changed.

Â It's at some new radius 2, which is also squared, and

Â it's now got some new velocity at this new radius.

Â So let's look at this equation we've come up with.

Â Notice that there's a mass that's the same on both sides,

Â the mass of the asteroid hasn't changed.

Â So that crosses out.

Â I also notice that I've got an r on either side.

Â They can kind of be cancelled out.

Â So what I'm left with is a very simple equation that's actually very powerful.

Â We use this or see this results come up quite often.

Â That if I know the distance and the linear speed of the asteroid at one place.

Â And maybe its location at another, I can calculate its speed,

Â which exactly what this question's asking us for.

Â So let me give myself just a little bit more room here.

Â It told me, for instance, at one point in space,

Â a distance of 3 times 10 to the 10th meters.

Â We knew the object was going 20 kilometers per second.

Â Which is 20 times 10 to the 3 meters per second.

Â It then tells me that its closest point.

Â Which is what we were looking for, a faster speed,

Â 2 times 10 to the 10th, distance in meters.

Â And I'm looking for the final speed.

Â So I can multiply the two terms on the left, divide by 2, times 10 to the 10th.

Â And for my final speed, a good check as usual,

Â make sure you're not going faster than the speed of light, 3 times 10 to the 8.

Â And we're nowhere near that in this problem,

Â 3.0 times 10 to the 4th meters per second.

Â Definitely faster than it was before,

Â like we said should happen when they're closer to the Sun.

Â A smaller radius and less than the speed of light.

Â Okay, so remember that momentum is a very helpful quantity when talking about

Â collisions and explosions.

Â And that's exactly what's going on in this problem.

Â I have a small ball of clay that strikes a larger sphere.

Â So here's my small ball of clay coming in and

Â it sticks to the sphere and this whole thing will start to rotate.

Â It wants to know,

Â what is the angular speed of that thing as it starts to rotate.

Â So, a couple of facts that they give us,

Â the speed is four meters per second at the beginning.

Â It also tells me that this m1 is a small one kilograms.

Â While this larger sphere, this m2, has a mass of three kilograms.

Â And it also give me the radius as 0.3 meters, so about 30 centimeters in radius.

Â Well since I know that this is a collision,

Â there's no external forces that are significant.

Â I know angular momentum before should equal angular momentum afterward.

Â And at the beginning of the problem,

Â there's only one object moving, this small mass.

Â And you might say to yourself, well, this first object is moving in a straight line,

Â it's not rotating.

Â You can't say that it has no zero, in this case, angular momentum.

Â An object moving in a straight line can have an angular momentum.

Â It depends on what you pick as the center, as the axis.

Â And in this case, this object starts to spin, clockwise.

Â If that's the case, right when they object reaches this point, and

Â it is moving with this speed, right before it collides with the sphere.

Â We're going to say that it has an angular momentum.

Â Because instantaneously,

Â it has a speed moving to the right which we can consider an angular velocity.

Â When we have a conversion between those two and

Â it's a certain distance from that radius.

Â So follow me here if you're a little lost and see if it makes more sense to you.

Â There's a moment of inertia of that point mass around the axis we've chosen and

Â it has an angular velocity.

Â So, I times omega.

Â And so, I'm going to call those for the first objects and their initial speeds.

Â Then after the problem, the two stick together.

Â So what have now is in a moment of inertia for

Â the sphere and the small mass that are stuck together.

Â So they have to have the same final angular velocity.

Â In fact,

Â it might even be easier then if I'm going to adjust my subscripts to match.

Â Let's make that an initial angular velocity.

Â This is a point mass at the beginning that's moving, so

Â I'm going to use m r squared.

Â So, it's an m1, r squared.

Â Now that r is, since it hits the edge of the disk or in this case sphere,

Â it's a distance r from the axis of rotation.

Â So that is the radius of the sphere that we're using there.

Â Again, I'm going to use our bridge equation.

Â V equals r omega, which comes up a lot in this unit.

Â So, v over r.

Â Now, after the collision, you have the point mass and the sphere moving together.

Â So, I need to sub those in.

Â Again, you have your mass 1 for your point mass, at a distance r squared,

Â because it's at the radius of the sphere plus two-fifths.

Â Now we have our mass of the sphere for the first time.

Â And again, the radius of the sphere and we're looking for

Â this final angular speed.

Â You use the same symbols here, so brackets.

Â I'm going to scroll down.

Â So let's take a look at what we have and what we don't have.

Â I know the mass of both, I know the radius of that sphere.

Â It looks to me like we're ready to sum the numbers and solve, so let's see how we do.

Â And we have a mass of 1 kilograms 0.3.

Â Now notice that the r's here can cancel out, so I only need to sub that in once.

Â I don't have to square it.

Â A speed of 4 meters per second on its way in, a linear speed.

Â Then on the right-hand side mass of 1 kilogram,

Â radius of 0.3 squared plus two-fifths.

Â The mass of the bigger sphere was 3, radius 0.3 squared and

Â then solving here for the final angular speed.

Â When I put all that into my calculator,

Â I divided by that huge term in the brackets to move it to the left-hand side.

Â I get a final angular velocity of 6.06 radians per second.

Â That's how fast the system is spinning after the collision.

Â Now we're not quite done, although we're getting there.

Â This question also asks how much energy was lost?

Â And typically in any real collision, energy is lost to sound,

Â to heat, especially an collisions like this.

Â So we have to calculate how much energy we had before the collision and

Â only one thing was moving.

Â We have the kinetic energy of that small point mass.

Â And you can do this as, in terms of rotation or in terms of linear.

Â It will come out the same.

Â I'm going to do linear, because we have those numbers and

Â I don't have to worry about using a bridged equation.

Â Just a one-half mv squared.

Â So one-half, 1 kilogram at speed of 4 squared.

Â Initially, at the beginning of this problem, we had 8 joules of energy.

Â Now, I need to find out how many energy do we have remaining in the problem

Â after the collision.

Â So for this, I think its a little easier to actually use rotations since we know

Â the final speeds.

Â So, after again, keeping track of my work carefully.

Â My total kinetic energy,

Â they will be one-half I omega squared plus one-half I omega squared.

Â Remember, we have the small point mass and the whole disk, that our sphere,

Â I'm sorry, that's actually rotating.

Â And this is a sphere that it hit, I apologize for mixing up the words.

Â So, one-half, now we're talking about that point mass, mr squared.

Â How far it is away from the center?

Â And we know the final angular speed, so that can just stay in.

Â Plus, since I'm not going to use a bridge equation one-half,

Â we're talking about the sphere, two-fifths mr squared.

Â And that's the mass of the big sphere this time.

Â So be careful not to mix them up.

Â Let's use different subscripts just to make sure and

Â then our final angular speed.

Â And they have the same final angular speed, because they're connected.

Â I can sub in my numbers, one-half the first mass.

Â Well, the mass of 1 kilogram.

Â The radius, 0.3 squared and

Â the final angular velocity that we solved,

Â 4.606 and one-half, two-fifths.

Â The mass this time of that larger object, the sphere, 3.

Â The radius again, 0.3 and

Â then the 6.06 squared, the final angular speed.

Â When I finally solved them for the total,

Â final kinetic energy, I got 3.636 joules.

Â Notice that we have lost energy,

Â it's definitely less energy than the 8 joules we started out with.

Â It specifically asked how much was lost.

Â So, if I want to know how much was lost,

Â I've got to subtract the 8 joules minus the 3.636 joules.

Â All right. Find that we lost 4.36

Â joules in this equation.

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