Main content

## Physics library

### Unit 7: Lesson 2

Torque, moments, and angular momentum- Introduction to torque
- Moments
- Moments (part 2)
- Finding torque for angled forces
- Torque
- Rotational version of Newton's second law
- More on moment of inertia
- Rotational inertia
- Rotational kinetic energy
- Rolling without slipping problems
- Angular momentum
- Constant angular momentum when no net torque
- Angular momentum of an extended object
- Ball hits rod angular momentum example
- Cross product and torque

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Angular momentum

AP.PHYS:

CHA‑4.D (EU)

, CHA‑4.D.2 (EK)

, CHA‑4.D.2.1 (LO)

, CON‑5.E (EU)

, CON‑5.E.1 (EK)

, CON‑5.E.1.1 (LO)

Introducing angular momentum conceptually starting from linear momentum. Also covers some real-life examples. Created by Sal Khan.

## Want to join the conversation?

- Is there a video that takes into consideration the moment of inertia.(22 votes)
- here in this video itself in the equation L=mw(r)square

m(r)square is the moment of inertia of the object(15 votes)

- I have a question :

If radius decreases then angular velocity or speed increases OK!

Now if something was travelling at a speed v1 before decrease in radius and v2 after decrease in radius then we know v2>v1 right if no force was applied to it.

so kinetic energy before was 1/2 mass v1^2

and after 1/2 mass v2^2

since v2>v1

kinetic energy after is greater than before.

So my question:

Where does the energy came from, isn't that breaking law of conservation of energy??

Thanks!!(2 votes)- Actually, the translational speed does not stay the same, considering that no torque was exerted on the system. What remains constant is the angular momentum, mvr. So, since the mass doesn't change, v.r is constant, thus decreasing the radius does increase the speed.

The extra kinetic energy comes from the force you exerted to pull the object from r1 to r2.

The net force exerted on the object at every moment, if we're decreasing the radius very slowly(we don't need to, its just to make it simpler), is mv²/r, the centripetal force.

So, the work done will be

W = -∫F.dr (from r1 to r2) =

-∫mv²dr/r =

-m.∫(v1.r1)².dr/r³ (from the condition v.r is constant, we have v.r =v1.r1 ---> v² = v1²r1²/r²)

= -m.(v1.r1)².[-1/2r2² + 1/2r1²]

= mv1².[(r1² - r2²)/r2²]/2

Which should be equal to the total change in energy.

ΔE = mv2²/2 - mv1²/2

= mv1²r1²/2r2² - mv1²./2 (here, I used v2r2 = v1r1)

= mv1².[(r1² - r2²)/r2²]/2

= W

So, all the extra energy comes from the work exerted by the pulling force.(19 votes)

- Does boomerang turns because of angular momentum?(4 votes)
- Yes, in a sense. It turns due to precession, an effect that causes many rather weird observations, like a spinning wheel with an axle and a pivot point. I recommend this video (and this youtube channel also) https://m.youtube.com/watch?v=ADJKsLEAOHo(2 votes)

- What is the difference between angular momentum and angular velocity(1 vote)
- let me ask you a question; do you know the difference between (linear) momentum and velocity?

Angular is virtually the same only it is rotational(5 votes)

- Sal said that L=mvr, where v is the magnitude of the perpendicular component of the velocity, but he also said L is a vector quantity. How is this so?(2 votes)
- Two questions: 1) Why don't we just have w NOT be in radians. Then all of our formulas could be simpeler. Maybe we could define it as orbital velocity or something? And 2) a read somewhere that orbital velocity doesn't actually limit to tangential velocity, so v does not equal rw. Is this accurate?(3 votes)
- The two questions are the same. The reason we use radians for angular velocity is because in that case, the unit radians falls straight out of the mathematics and r*w, when we write angular velocity in radians, is equal to v. If we were to write angular velocity in rotations per unit time rather than radians per unit time, then they would not be the same.(0 votes)

- what is tangential velocity and why it is perpendicular to radius in above video?(0 votes)
- To elaborate on that, any 2D vector can be broken down into 1D components of that vecotor. Typically the vectors we choose to use are vectors that are perpendicular to each other because that makes things simplest. For instance, a vector pointing North-East can be considered as the sum of two equal magnitude vectors, one pointing North and the other East.

When considering circular motion, it's often easier to decompose vectors into tangential (perpendicular to the radius) and radial (along the direction of the radius). This is because if the components were defined in terms of cartesian coordinates (e.g. X and Y or North/South and East/West) the velocity of a spinning object would be constantly changing as the object moves around the circle. By discussing tangential and radial velocity, the analysis becomes much simpler. (typically, tangential velocity is constant and radial velocity is zero).(7 votes)

- At4:12, perhaps they use "L" because it resembles the sign "∠" used to represent angles.

http://www.rapidtables.com/math/symbols/Geometry_Symbols.htm(2 votes)- Nice one:

or maybe as sal pointed out:

- A was already used for "Area/magnetic vector potential/amplitude".

- a for acceleration.

- M for "moment of force/magnetization".

And L was simply used because of the sound of the latin word "Angulus".

or:

The scientist who is responsible for the naming just looked out the window and the first thing he saw started with the letter L.

We will probably never know.(36 votes)

- Around1:40, Sal mentions that if there is no impulse or net force, then there will be no change in momentum. That much makes sense. But Sal also mentioned that without impulse or net force, there is a conservation of momentum.

Isn't momentum always conserved in a system, especially if it is an isolated system? That is, the momentum that is lost by one object is gained by another?(1 vote)- momentum is always conserved in a closed system

If you apply an impulse from outside the system then the momentum in the system will change

If you take a broader view of the system to include that source of impulse, then still momentum was conserved.(3 votes)

- BUt why when skater turn we do a calculation with momentum, There is no crash, should not we calculate on kinetic energy?(2 votes)

## Video transcript

If we have some mass, m, and it is moving with some velocity, let's say the magnitude of
that velocity we say is v, we know that this object
right over here has momentum. Translational momentum. And that momentum, and we
use the Greek letter rho to represent momentum. Translational momentum
is defined as being equal to the mass times the velocity. This is all a review, we have other videos where we talk about
translationial momentum, and one way to think about it is, "Well how hard is it to stop this thing?" Literally in everyday language you think, "Well how much momentum
does something have? "The more momentum this has
the harder it's going to, "the harder it is to stop it in some way." And so we know if we wanna get a little bit more mathematical, that if we wanna change momentum, we have to apply force
for some amount of time. And so the magnitude of our force times the duration of the
time that we apply it for, force times time and
this is called impulse. And this is once again, all review. This is equal to change in momentum. Let me do this in that yellow color. That is equal to change in momentum. So if you don't have any impulse, especially if you don't have any net force acting on an object, its momentum is going to be constant. You have a conservation of momentum. And we use that idea in all sorts of interesting physics
applications in the world, and especially a lot of
cases using billiard balls and whatever else. So now let's try to take a similar idea, but go into the rotational world. So let's imagine you have a mass, for the sake of this we're
gonna assume it's a point mass. So you have a mass there. And let's just say it's attached by, essentially a massless wire, to you know, it's just nailed down right over here. And so this right over here would be its center of rotation, and so you could imagine
if someone applied a torque to this mass, this mass could start
rotating in a circle. And you can just assume that maybe it's sitting on a, you know the screen, this is kind of a frictionless surface, there's no air resistance. And so then it will, if
you apply a torque here, it will start rotating. And so you could think about, "Well there might be an idea, "just as momentum is this idea of, "well how hard is it to stop something?" You might say, "Well how?" And this is, stop translating
something from moving. You might think, "Well
maybe there's a similar idea "of how hard is it for something, "or how hard is it to
stop rotating something?" And you could imagine that
that idea has been defined and it has been defined
as angular momentum. So let me make this clear, this right over here is momentum. And over here we'll talk
about angular momentum. And actually both momentum
and angular momentum are vector quantities. So here I just wrote
kind of the magnitudes of velocity and momentum. But momentum is a vector
and it could be defined, the momentum vector could be
defined as equal to the mass which is a scalar quantity
times the velocty. Times the velocity vector. Now the same thing is
true for angular momentum, but I'm gonna stay focused on the magnitude of angular momentum. Angular momentum can have
direction as you can imagine you could rotate in two different ways, but that gets a little
bit more complicated when you start thinking about
taking the products of vectors because as you may already know or you may see in the future, there's different ways of
taking products of vectors. But just to get the intuition
of angular momentum, I'll focus on the magnitudes. So angular momentum is defined and the letter used is L. I did a lot of research
to try to figure out why it is called L, and I
could not find a good reason. So in the message board below
if anyone has a good reason I would like to know why angular momentum is called L. A lot of the best arguments I saw is that almost everything else was, all the other letters were used up for other ideas in physics. But anyway, angular momentum is defined, and it's defined very similarly. Just as kind of torque
is the thing that can change how something rotates, and force is the way
that something changes how something translates it, and torque is force times distance from the center of rotation, everything in kind of the rotational world is defined in a similar way. You kind of take the analogue
in the translational world, and you multiply it times the distance from your center of rotation. So angular momentum is defined as mass times velocity times distance from the center of rotation so let's call this distance
right over here, r. r for radius 'cause you could imagine if this was traveling in a circle that would be the radius of the circle. m, v, r. Actually let me be a little
bit more careful here. It's the magnitude of the velocity that is perpendicular to the radius. Sometimes it might be called
the tangential velocity. So this symbol right over here, this is the magnitude of the velocity that is perpendicular to the radius. So it would be that
magnitude right over here. This is what we define
as angular momentum. Now what I will tell you here is, just as in the absence of a net force, momentum is constant. We know, and I haven't
shown it to you yet, I haven't proven it to
you yet mathematically, but in the absence of torque, so if torque is equal to zero, we'll do torque in pink. If torque is equal to zero, if there's no net torque going on here, if the magnitude of
torque is equal to zero, then we will have no change. No change in angular momentum. And we will look at that
mathematically in a few seconds. But just from this there's a very interesting
thing that arises. And something that you might have observed at even the Olympics or in other things. And this is the idea that you can, by changing your radius, you could actually change
your tangential velocity. And as we've seen in previous videos, tangential velocity is closely related to angular velocity. So let's explore that a little bit. So when we write it in the world where, well actually you see
it straight out of this, if L is constant, if r went down, so let me write this down. So let me rewrite it over here. So L, whoops. L is equal to mass times
tangential velocity, or actually well yeah,
tangential velocity, or the velocity that's
perpendicular to the radius, times the radius. Now what happens, if we assume that this is constant, if we assume that there's no torque, so we're in this world. So this over here is going to be constant. So what happens if we were to reduce r? Somehow this wire started
to reel in a little bit or started to wrap around here, and that's actually a reasonable thing, you could imagine as it rotates
it starts to wrap around this thing so the wire gets shorter. So if r goes down, and this is constant, the
mass isn't going to change, Well if L is constant,
mass isn't changing, r is going down, tangential velocity, or the velocity that's
perpendicular to the radius is going to go up. And if we wanna think about it, we can think about it in
terms of angular velocity, we know that angular velocity, which we would measure
in radians per second, we would use the symbol omega, omega is defined, and we go into much more
depth in this in other videos, as tangential velocity, the magnitude of the velocity that is perpendicular to the radius, divided by the radius. Or if you solve for tangential velocity, you get v is equal to is equal to omega r. And so if you substitute back into this, really this definition
for angular momentum, you get angular momentum is equal to mass times this times r. So mass times, I'm just
substituting for velocity here, times omega r, times r. Which of course is just omega r squared. So once again we do the same exercise. If our radius goes down, what happens to our angular velocity? Remember we could measure this in angles per second or radians per second, well if this is constant, remember we're assuming
that there's no net torque being applied to the system, so we're still in this
world right over here. If we assume that this
thing isn't changing, but the radius were to change, what's going to happen to omega? Well omega is going to go, omega is going to go up. Now likewise, if the radius got longer, So the radius got longer, what's going to happen to omega? Omega is going to go, omega is going to go down. So if you reduce the radius you're going to start spinning faster if you increase that radius, you're gonna start spinning slower. And you have seen this, or I think there's a high likelihood that you have seen this probably at the Olympics when
you have seen figure skaters. Where they might start spinning and they have their arms out. So when their arms are out you could say that their
radius is further out. And obviously a figure skater is a much more complex
system than a point mass. You could imagine a figure skater is a bunch of point masses. Well you could just model a figure skater as a huge number of point masses at different radii, and you would wanna sum
up their angular momentum, but the essence of what is happening is is when her arm is out the average radius when you're calculating all of the point masses in
her arms and all the rest, the average radius is higher. And then when she pulls them in, when she pulls them in
that radius goes down, and her angular velocity goes up. And you see that. They start spinning, and then without applying any torque, when they pull their arms in they start spinning faster. And then if they push their arms out, once again without applying any torque, they start spinning slower.