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Lecture Comments (6)

1 answer

Last reply by: Professor Dan Fullerton
Sun May 10, 2015 3:33 PM

Post by Nitin Prasad on May 10, 2015

On the 2014 frq part b, why is the final momentum 0?

1 answer

Last reply by: Professor Dan Fullerton
Thu Apr 23, 2015 10:41 AM

Post by Micheal Bingham on April 23, 2015

For part (a) of the 2005 FR3, why do we use the equation for spin angular momentum even though the object is not rotating about its center of mass?

1 answer

Last reply by: Professor Dan Fullerton
Fri Apr 17, 2015 6:05 AM

Post by Mus Kastrati on April 16, 2015

Hi dan, can you explain why you distribute m to r and v.with regards to the cross product.

Angular Momentum

  • Momentum is a vector which describes how difficult it is to stop the translational motion of an object.
  • Angular momentum is a vector which describes how difficult it is to stop the rotational motion of an object.
  • For an object rotating about its center of mass, the spin angular momentum of the object is constant, regardless of reference point.
  • Spin angular momentum, the product of an object’s moment of inertia and its angular velocity about the center of mass, is conserved in a closed system with no external net torques applied.
  • A torque on an object change’s the object’s angular momentum.

Angular Momentum

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 0:09
  • Linear Momentum 0:44
    • Definition of Linear Momentum
    • Total Angular Momentum
    • p = mv
  • Angular Momentum 1:08
    • Definition of Angular Momentum
    • Total Angular Momentum
    • A Mass with Velocity v Moving at Some Position r
  • Calculating Angular Momentum 1:44
    • Calculating Angular Momentum
  • Spin Angular Momentum 4:17
    • Spin Angular Momentum
  • Example I: Object in Circular Orbit 4:51
  • Example II: Angular Momentum of a Point Particle 6:34
  • Angular Momentum and Net Torque 9:03
    • Angular Momentum and Net Torque
  • Conservation of Angular Momentum 11:53
    • Conservation of Angular Momentum
  • Example III: Ice Skater Problem 12:20
  • Example IV: Combining Spinning Discs 13:52
  • Example V: Catching While Rotating 15:13
  • Example VI: Changes in Angular Momentum 16:47
  • Example VII: AP-C 2005 FR3 17:37
    • Example VII: Part A
    • Example VII: Part B
    • Example VII: Part C
    • Example VII: Part D
  • Example VIII: AP-C 2014 FR3 24:23
    • Example VIII: Part A
    • Example VIII: Part B
    • Example VIII: Part C
    • Example VIII: Part D
    • Example VIII: Part E

Transcription: Angular Momentum

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

I’m Dan Fullerton and in this lesson we are going to talk about angular momentum and conservation of angular momentum.0003

Our objectives include calculating the angular momentum vector for moving particle.0010

Calculating the angular momentum vector for rotating rigid object or angular momentum is parallel to the angular velocity.0015

Recognizing conditions under which angular momentum is conserved.0023

Stating the relationship between net torque and angular momentum.0027

Analyzing problems in which the moment of inertia changes as an object rotates.0031

Finally, analyzing collisions between moving particles and rigid rotating objects.0036

Let us take a look by starting off with a quick review of linear momentum.0042

Linear momentum with the symbol P is a vector describing how difficult it is to stop a moving object.0047

The total momentum is the sum of individual momentum when you are dealing with the system comprised of smaller objects.0053

A mass with velocity V has momentum P equal the MV, our formula for linear momentum.0060

Looking in the rotational world, we have a rotational analog linear momentum and its angular momentum.0069

Angular momentum given the symbol capital L is a vector describing how difficult it is to stop a rotating object.0074

Just like with the linear momentum, the total angular momentum is the sum of the individual angular momentum when you have a more complex object.0082

A mass with velocity V, moving at some position R about 0.2 has angular momentum L with respect to q, the angular momentum about point q.0090

Our calculation of angular momentum is going to get a little bit more complicated.0100

Calculating angular momentum, let us assume that we have some mass M situated at distance from point q, we will define an R vector, 0107

position vector from q our reference point to our mass M.0116

Our mass M is moving with some velocity V in this direction which means its momentum, its linear momentum must also be in that direction.0121

The angle between Rand V we will define as θ.0130

By definition, the angular momentum about point q is equal to our position vector from our reference to our object crossed with its momentum vector, 0134

its linear momentum, which is going to be R crossed with MV.0144

Our definition of linear momentum.0148

Mass is a constant so we can pull it out and get our cross V × the mass.0151

The direction of angular momentum is given by the right hand rule.0158

Let us make sure we specify that.0161

Direction by right hand rule because it is a cross product.0163

Take the right hand, point the fingers of your right hand in the direction of the position vector, bend them in the direction of the velocity of the particle 0168

and you will have your thumb pointing in the direction of the angular momentum.0176

In the case of this diagram here, our angular momentum vector is going to be into the plane of the screen.0181

If we wanted to look at the magnitude of the angular momentum about point q, we can say that that is going to be MVR sin θ.0189

We also know, if we start to look at something like an object that is rotating in a circle, let us make that point q,0203

we will define our position vector up that way with an object here moving with the velocity there.0210

Our angle is 90°, if that is the case, and we also know that V = ω R for something traveling in a circular path,0216

we can say the magnitude of the angular momentum is going to be M ω R².0226

Or if we rearrange that a little bit, the magnitude of the angular momentum vector about point q is MR² ω.0235

Note that MR², how that is starting to look like a moment of inertia, we are not exactly there yet but 0246

you are starting to see a relationship that is coming out as we define angular momentum.0252

For an object rotating about its center of mass, the angular momentum is equal to the moment of inertia × the angular velocity.0259

That is known as an object's spin in angular momentum.0267

Spin angular momentum is constant regardless of your reference point, regardless of what point you pick about which you are measuring the angular momentum.0271

Regular angular momentum depends on your reference points.0280

Spin angular momentum, when it is rotating about its center of mass does not.0284

Let us take a look at a couple examples here to get started.0291

Find the angular momentum of a planet orbiting the sun assuming that perfectly circular orbit.0294

Let us define our planet and make a couple of different positions.0300

If we started over here with some tangential velocity V, we will call that V1 that position 1 and 0304

maybe define a vector R1 from our reference point our center point we will call the q.0312

There is our R1 vector, our angle here must be 90°, it is moving with some angular velocity ω.0319

In a later point in time it is over here.0327

We will define our vector R2 to our mass moving with velocity V2.0330

Our angle there is 90°, our angular momentum about point q is R crossed P which is going to be R crossed with MV or R crossed with V × M.0341

The magnitude of that is just going to be MVR sin θ but since θ = 90° we can write that as MVR and 0362

the direction again point the fingers in the direction of the position vector and bend in the direction of velocity, 0374

I have an angular momentum vector pointing into the screen again.0381

It is a pretty straightforward, when the angular momentum of a planet orbiting the sun MVR into the plane.0386

How about some particles?0395

Find the angular momentum for 5 kg point particle located at 2, 2.0397

There is a particle with the velocity of 2 m/s east and first we are going to find it about the origin, about 0, 0.0401

Let us start there, the angular momentum and we will define the magnitude of it.0410

The magnitude of the angular momentum about the origin is going to be MVR sin θ.0415

Our mass is 5 kg, our velocity 2 m/s east, our distance, our position vector from 0 to there, if that is 2 and that is 2, 0424

by the Pythagorean Theorem that is going to be 2√2 and sin θ is going to be sin 45° which is √2/2.0436

That is going to be 20 kg m²/s.0447

Let us find it about point P over here at 2, 0.0453

The angular momentum to about point P same formula MV R sin θ, same mass 5, same velocity 2.0458

The position vector, the magnitude of R is just 2 this time, sin θ all that is going to be sin 90°, the angle between our position vector and the velocity vector.0471

sin 90° is just 1, 5 × 2 × 2 is going to be 20 kg m²/w.0482

Let us find the magnitude of the angular momentum about 0.2 here at 0, 2.0492

As I look at that, the magnitude of the angular momentum about point q MV R sin θ which is going to be 5 × our velocity 2 × R again, 2 × the sin of, 0500

Now, our position vector and our velocity vector, angle between them is 0.0519

That is going to be sin 0, sin of 0 is just 0 so our angular momentum about point q is 0.0524

Notice, how we have a changing angular momentum depending upon our reference point.0536

Let us take a look at angular momentum and net torque.0543

If we start off with angular momentum, our definition is the position vector crossed with its momentum and we take the derivative of both sides.0547

We will have to remember how we do the derivative of a cross product.0559

The derivative with respect to T of A cross B is going to be the derivative of A with respect to T crossed with B + A crossed with the derivative of B with respect to T.0563

So that implies if we take the derivative of both sides with respect to time, on the left hand side I have the derivative of my angular momentum 0582

with respect to q must be equal to, we will take DR DT cross with momentum + R crossed with DP DT.0590

If you recalled DR DT, this is what we call velocity.0609

The derivative of momentum with respect to time, that, we know as force.0616

We can write this as the derivative of the angular momentum with respect about point q with respect to T is equal to 0624

we will have our velocity V crossed B + our position vector cross with F.0634

Another simplification here, V cross P, the velocity and the momentum vector are in the same direction.0648

Their cross product is going to be 0.0656

Since, we know V crossed P= 0, determine that the derivative, the angular momentum about point q with respect to T = R cross F.0658

R cross F should look familiar though.0675

R cross F was our definition of torque.0678

We could write then that the derivative of the angular momentum about point q with respect to T is going to be equal to the torque about point q.0684

What does that mean?0697

A torque on an object is going to change the objects angular momentum or if you have a change in angular momentum that must be caused by some torque.0698

Torque change angular momentum.0708

Onto the conservation of angular momentum and other conservation law.0714

Spin angular momentum, the product of an object's moment of inertia and 0718

angular velocity about the center of mass is conserved in the close system with no external net torque applied.0722

Spin angular momentum Iω remains constant unless you have an external torque.0730

We can use that in many situations.0735

We are going to look at one of them right now.0738

An ice skater spins with a specific angular velocity.0742

She brings her arms and legs closer to her body reducing her moment of inertia that half its original value.0745

You have probably seen that if you watch a figure skater spinning.0751

They have their arms out, they bring them in, and they start spinning faster.0754

What happens to her angular velocity and what happens to her rotational kinetic energy?0758

We know angular momentum is Iω and that has to remain constant if we do not have external torque and we do not.0763

If we bring moment of inertia down, if we make that half, and angular momentum must stay the same, we have to double the angular velocity.0772

Angular velocity doubles.0781

What happens to rotational kinetic energy?0785

Rotational kinetic energy is ½ I ω².0788

If moment of inertia is cut in half but we double angular velocity and it is squared, we are going to end up with double the kinetic energy.0795

Angular velocity is doubled, rotational kinetic energy is doubled.0810

Where the extra energy come from?0814

She had some kinetic energy K knot, she got twice that amount.0818

The skater must do work to bring arms and legs and reduce that moment of inertia.0822

That work becomes the kinetic energy of the skater.0826

Taking a look at another example, we have a disk with moment of inertia 1 kg m² spinning about 0833

an axle through its center of mass with an angular velocity of 10 radians/s.0838

An identical disk here on the right which is not rotating is sitting along the axle until it makes contact with the first disk.0843

If the 2 disks stick together what is their combined angular velocity?0851

We could take a look at this from conservation angular momentum that the initial angular momentum must equal the final angular momentum.0856

Or initial moment of inertia × initial rotational velocity must equal final moment of inertia × final angular velocity.0865

If I solve for final angular velocity that is going to be I initial ω initial /I final.0875

Our initial moment of inertial was 1 kg m², our initial angular velocity was 10 radians/s, and our final moment of inertia, 0886

if the moment of inertia of one of these is 1, the moment of inertia of two of these must be 2, so 2 kg m².0897

Therefore, ω final must equal 5 radians/s.0906

Let us take another example, Angelina spins on a rotating pedestal with an angular velocity of 8 radiance/s, 0915

Bob throws an exercise ball which increases her moment of inertia from 2 kg m² to 2.5 kg m².0922

What is our regular velocity after catching the exercise ball and we are going to neglect any extra net torque from the ball’s forces 0930

caused by the ball and assuming magically she catches the ball and her moment of inertia goes up without having any other outside effects.0937

Since, there is no net external torque we know the initial spin angular momentum must equal the final spin angular momentum.0946

We can solve for Angelina’s final angular velocity, L initial = L final which implies that I initial ω initial, let us write the final to make it easier, equal I final ω final.0953

Ω final again = I initial ω initial/ the final moment of inertia.0972

Initial moment of inertia was 2 kg m², initial angular velocity was 8 radiance/s and final moment of inertia is 2.5 kg m² after she catches the ball.0980

That is going to be 4/5 of 8 or 6.4 radians/s.0994

A constant force F is applied for a constant time of various points of the object below.1009

Write the magnitude of the change in the object’s angular momentum due to the force from smallest to largest.1014

Remember, its torque that changes angular momentum.1020

We need to really look at this in terms of the torque and once we know that in terms of torque this becomes a pretty easy ranking test.1023

Our smallest torque, of course, is going to be from B, next we will have C and then we will have A and then we will have D, 1031

the most force applied the furthest distance and at the closest angle to that center of mass line.1040

B, C, A, D would be the ranking of the objects angular momentum due to the force from smallest to largest.1046

Alright, let us finish up with a couple of old AP free response problems.1056

We will start off with a 2005 exam Mechanics question number 3.1061

You can find it here, take a minute to download it, look it over, give it a try, and come back here and see if we can do it together.1065

Alright, in this problem we have a ball that is at rest to the uniform rod that is swinging in the pivot point initially rotating at an angular speed ω.1074

And it looks like we are looking initially for the angular momentum of the rod about point P before they collide.1086

The angular momentum of the rod about point P is just its moment of inertia × its angular velocity.1094

It tells us the moment of inertia of the rod is M1 D² /3 ω.1101

There is our answer.1109

Moving on to part B, derive an expression for the speed of the ball after the collision.1114

Here we can use conservation of angular momentum.1119

A moment of inertia of the ball about point P must equal the moment of inertia of the rod about point B.1122

This is while the ball is moving, this while the rod is moving, which implies then that the moment of inertia or the angular momentum of our rod, 1130

we just figured out was M1 D² /3 ω and that must be equal to the angular momentum of the ball about point P, is going to be its mass × its velocity × its distance.1139

We do not have a sin θ there because that is going to be 90° according to our diagram.1155

Therefore, all we have to do is solve for V.1160

I find that V is going to be equal to, we have M1 D² /3 ω.1162

We have got a D down here and then M2, which implies then that our V is going to be M1 D ω /3 M2.1174

Alright there is B, let us give ourselves a little more room here for part C.1191

C, says, assuming that this collision is elastic, find the numerical value of the ratio of M1/M2.1198

If it is elastic collision that means the total kinetic energy before the collision must equal the total kinetic energy after the collision.1207

Kinetic energy initial = kinetic energy final which implies that the kinetic energy of the rod before the collision must equal the kinetic energy of the ball after the collision.1216

Or kinetic energy of the rod ½ I ω² must equal ½ M2 V², which implies then looking at our rods kinetic energy that is going to be M1 D² ω² /2 × 3 = ½ M2 V²,1228

which is going to be M2 M1² D² ω² / 2 × 9 M2².1253

Let me square our V from our previous answer.1264

As I look to see what I can simplify here, we have got M1 here, we have got an M1² there, we have got an M2², we can get M2.1268

We got 2 × 3 vs. 18 over here.1278

This can be 6, that will be 18 D² ω².1285

I get that 1/6 = M1/18 M2 or M1/M2 must equal 3.1294

Alright, let us take a look here at D, a new ball with the same mass M1 as the rod is now placed at distant x from the pivot, 1312

assuming the collision is elastic for what value of x will the rod stop moving after hitting the ball?1321

This looks like a conservation of angular momentum problem again.1328

Angular momentum of the ball at point P = angular momentum of the rod at P so we had M1 D² /3 ω = M1 Vx.1332

Therefore, the velocity of the ball is going to be D² /3x ω.1347

When we go to look at our kinetic energy, we have kinetic energy of the rod before the collision must equal 1354

the kinetic energy of the ball after the collision because it tells us it is elastic,1362

which implies that ½ I ω² = ½ M1 V², which implies that, let us see, we have ½ M1 D² /3 ω² for our left hand side.1367

In the right hand side, we have ½ M1 and now V² is going to be D⁴/9 x² ω²,1386

which implies then that M1 D² ω² /3, factor out the 1/2 must equal, we will have M1 D⁴ ω²/9x²,1398

which implies then as we do a little bit of simplification M1 M1, that will be D² ω² ω², 3 in the 9, 13.1417

I get that we have 1= D² /3x² which implies the net x² is going to be equal to D² /3 or x = D/ √3.1433

That finishes up that problem.1457

Let us take a look at one more.1460

Let us go to the 2014 exam free response number 3.1464

I will give you a minute to get that printed out and look it over.1468

In this problem, we have got a large circular disk of mass M, radius R, a mass of M/2, standing on the edge of the disk.1472

It got a large stone mass M/20, throws that stone horizontally, how long will it take the stone to strike the ice?1480

This is a kinematics problem, we have been doing this problem for quite a while now.1488

We called down the positive y direction from vertically the initial is 0, Δ with y is that height H, acceleration in the y is GD acceleration due to gravity.1493

Δ y = V initial T + ½ AYT², V initial is 0.1506

This implies then that our Δ y is H is going to be ½ GT² or T is going to be equal to √2 H/G.1516

Part B, assuming the disk is free to slide on the ice, find an expression for the speed of the disk and person after the stone is thrown.1534

That looks like a conservation of momentum problem, an explosion in one dimension.1541

For part B, our initial momentum must equal our final momentum.1546

Our initial momentum, our mass is M + M/2, the man + the disk × V + M/20, our stone × V initial, all of that is going to equal 0, that is our final momentum,1552

which implies then that 3 M/2 V + M/20 V 0 = 0.1568

In just a little bit of algebra here, 30 MV + MV initial =0.1578

Therefore, 30 V + V0 = 0 which implies then that V0 is just going to be equal to V=-V0 /30.1586

We have our expression for the speed of the disk and person after the stone is thrown.1604

Let us take a look now at part C, let us give ourselves some more room here.1613

For part C, derive an expression for the time it will take the disk to stop sliding.1619

I’m going to start with a free body diagram for the disk and then we have got normal force, 1624

we have got its weight which is going to be 3 M /2 G and the force of kinetic friction.1629

Net force in the x direction is going to be -FK which is -friction is fun, μ × the normal force which is –μ.1638

Our normal force is going to be 3M /2 G and all of that has to equal our total mass 3 M/ 2 × the acceleration in the x direction,1648

which implies then, 3 M /2 we can divide out and find out that ax = -μ G.1660

If we want to know how long it is going to take to stop sliding, we can go to our kinematics V final = V initial + acceleration × time 1669

or time is going to be our V final – V initial/ acceleration which is 0 - V initial V knot /30 ÷ μ G.1682

Complies then that our time is just going to be V0 /30 μ G.1697

Moving on to part B, it looks like we have got a fixed poll through the center of our disk so it can rotate on the ice.1709

The person throws the same stone horizontally, tangential direction, initial speed, and a rotational inertia of the disk is MR²/2.1721

Find the angular speed of the disk after the stone is thrown.1730

For part D, the moment of inertia of the disk is ½ MR²,R².1736

We have got the moment of inertia of the man on the edge of the disk and that is going to be M/2 R², his mass × the distance from that center point.1747

That the total moment of inertia is going to be, we have got ½ MR² + ½ MR².1760

The man + the disk is going to be MR².1766

We can use conservation of angular momentum, the initial angular momentum of our system must equal the final angular momentum of our system, 1770

which implies that the moment of inertia × the initial angular velocity must equal the moment of inertia × the final angular velocity.1779

Initially, our angular momentum is 0 so that has to equal our total after, our stone M /20 V knot R + we have our moment of inertia of our man disk system MR² × ω.1789

Solving this for ω, ω = MV knot R/ 20 MR².1810

It looks like I’m not going to worry about magnitudes here.1822

M will make a ratio of 1, we will have 1 /R, ω = V knot /20 R .1827

Alright, that looks good for D.1843

Now for part E, the person now stands on the disk at rest R/ 2 from the center of the disk, halfway to the outside.1846

Throws a stone horizontally with speed V knot again, in the same direction, what happens to the angular speed of the disk after throwing the stone?1856

Let us see, for part E, our moment of inertia of the man disk system is changing again.1865

It is going to be ½ MR², the moment of inertia of the disk but the moment of inertia of the man now is his mass M/2 × the square of his distance from the center which is R/2².1870

That is going to be ½ MR² + MR² /8.1883

4/8 + 1/8 that is going to be 5 MR² /8 for the new moment of inertia.1891

Our moment of inertia is going down.1897

Let us apply conservation and angular momentum again and see what happens.1901

L initial must equal L final, which implies that I ω initial = I ω final or again 0 = M /20 V knot, now our distance here is R/2 + 1905

our moment of inertia 5 MR² / 8 × our new angular velocity ω.1927

A little bit of math here, MV knot /40 = 5 MR/8 ω dividing by one of those R, complies then that ω is going to be equal 8 V knot /200 R.1935

That is going to be 4/102 or 50/125 V knot /25 R.1956

That went down so it has to be less than.1963

Hopefully, that gets you a pretty good start in understanding of angular momentum and conservation of angular momentum.1971

Thank you for joining us here on www.educator.com.1977

I look forward to seeing you again real soon and make it a great day everybody.1979