For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### 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
- Objectives
- Linear Momentum
- Angular Momentum
- Definition of Angular Momentum
- Total Angular Momentum
- A Mass with Velocity v Moving at Some Position r
- Calculating Angular Momentum
- Spin Angular Momentum
- Example I: Object in Circular Orbit
- Example II: Angular Momentum of a Point Particle
- Angular Momentum and Net Torque
- Conservation of Angular Momentum
- Example III: Ice Skater Problem
- Example IV: Combining Spinning Discs
- Example V: Catching While Rotating
- Example VI: Changes in Angular Momentum
- Example VII: AP-C 2005 FR3
- Example VIII: AP-C 2014 FR3

- 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

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### 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

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.