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# Angular momentum of an extended object

In this video David derives and shows how to use the formula for the angular momentum of an extended object. Created by David SantoPietro.

## Want to join the conversation?

• Angular momentum of a rod rotating about one end: David did not complete the calculation for momentum because he didn't explain how m(rsqr'd) became 1/3m(rsqr'd). Was he trying to avoid the use of integrals?
(8 votes)
• Short Answer - Yes he was trying to avoid the use of integrals and you can choose to just trust him or...

Long Answer - I answered a similar question to this on another Moments, Torque and Angular Momentum Video

We know that the total moment of inertia for a set of point masses is the sum of each of their masses multiplied by the square of their corresponding distance from the pivot point.
Total Moment of Inertia = Σmr²
We can consider a rod of uniform density as an infinite number of infinitesimally small point masses and thus find a similar sum to the one above. We can represent this as a definite integral. If we take the moment of inertia to be from the end of a rod of length L and total mass M we can calculate this moment of inertia.
Total Moment of Inertia = (From 0 to M)∫ r² dm
This effectively takes the sum of an infinite number of point masses in the rod multiplied by the square of their distances from the pivot point. However we can't integrate r² with respect to m. We can use the formula
m = (r/L)M
r/L represents the ratio of the current distance from the pivot point to the total length of the rod. We can multiply this by the total mass of the rod M to give us the total mass of the rod up to the distance r away from the pivot. This formula only works because we assume that the rod is of a uniform density.
m=(r/L)M = (M/L)r Now we can differentiate this equation with respect to r
dm/dr = M/L
Total Moment of Inertia = (From 0 to M)∫ r² dm
Now using the substitution dm = (M/L)dr and changing the limits of our integral accordingly our Total Moment of Inertia becomes
(From 0 to L)∫ (M/L)r² dr = [(M/L)(1/3)r³] (From 0 to L) = (M/L)(1/3)(L)³ - (M/L)(1/3)(0)³ therefore
Total Moment of Inertia = (1/3)ML² I hope that explains it and I didn't over complicate things!
(21 votes)
• At ,
when the units were multiplied ,where did the radian go.
ie:kgX(m^2)X(rad/sec),it became
kgX(m^2)/sec
(7 votes)
• Where did the 1/3 come from when doing the example?
(6 votes)
• from parallel axis theorem, we got Icom+Mh^2 = (1/2ML^2)+(M(L/2)^2) = 1/3ML^2
(1 vote)
• So, if you're rolling coins into one of those hyperbolic funnels that are used for donations. What physics and associated mathematics could you use to determine what coins would go down the funnel faster? I'm just stumped as to how you could apply this content to that situation, but I'd assume it's similar to orbits, but with a component of rotating disks.
(6 votes)
• Can angular momentum be conserved about any other point inside or outside the body, other than the center of mass of a body??
(3 votes)
• angular momentum is always conserved, regardless of the origin of your reference frame
(4 votes)
• Is the angular velocity constant, because all the small infinitesimal masses rotate through the same angle theta , in the same period of time?
(4 votes)
• correct
(1 vote)
• Guys please help, perhaps it is a ridiculous thought, but it just came to my mind that we may consider a ball of mass, traveling along a straight line as rotating around an infinitely large circle. In such case R would be infinite. Does it mean that a translating ball has an infinitely large angular momentum! So please point out where is a mistake here.
(2 votes)
• my intuition goes to the similar direction as yours. and that says yes

1. if R gets close to infinity, angular momentum goes to infinity too. but what's more interesting is angular velocity as well as acceleration goes to 0 in such a case. in other words, the mass would move in a "straight" line perpendicular to the invisible bar connecting it and its orbital center of mass

2. so i think it's not because it travels along a straight line thus it has an infinite R. but the other way around

3. if the distance of a body from the orbital centre goes to infinity, it has an incredibly small (if not at all) angular velocity and acceleartion. thus it can have a straight travel

and the opposite edge seems also interesting. if a body get closer to the orbital centre, say R=infinitely small or ~0, then its angular velocity and acceleration grows so big that it has only one motion, spin on the same spot
(2 votes)
• I'm confusing that how can we simplified the R1,R2,R3....Rn into just a single R in the eventual function
(2 votes)
• Because all the R1R2R3R4Rn are used in the equation as the summation of all the radii. That is how all the Rs were able to be just an summation R and eventually the moment of inertia
(1 vote)
• A rod of mass m and length l whose linear mass density increases linearly from zero at a end is hinged to the ceiling at the less denser end. Initially it is held horizontal and released.What is its angular acceleration immediately after release.
(0 votes)
• where does the 1/3 come from in the "L=(1/3mL^2)w?
(3 votes)
• The length in the moment of inertia is calculated as the total length of the object or the maximum distance of the object from the pivot point. For example if the pivot point of the rod was at its centre instead of at one end, how would you calculate the moment of inertia then?
(1 vote)

## Video transcript

- [Voiceover] So we saw in previous videos that a ball of mass m rotating in a circle of radius r at a speed v has what we call angular momentum, and the symbol we use for angular momentum is a capital L, and the amount of angular momentum that it would have would be the mass of the ball times the speed of the ball, so that means this is basically just the magnitude of the momentum, but then we multiply by the radius of the circle it's traveling in, and that gives us the angular momentum of this ball going in a circle, which is great and good to know, but sometimes you don't have a ball going in a circle and you wanna know the angular momentum. So, for instance, instead of this case, let's say you have this case, where instead of a ball going around in a circle, you've got a rod of mass m and radius R and the whole rod rotates around in a circle. Let's say the outside edge travels at a speed v just like the ball did. So the question is, will this rod also have an angular momentum that's equal to mvR and it won't. You can probably convince yourself of that because for the ball all the mass was traveling at a speed v and all the mass was at the outside edge of the circle that it traces out. In other words, all this mass is traveling at a radius of R. But for this rod, some of the mass, in fact, only part of the mass, only this outside edge of the mass, is actually traveling at a radius R. That's the part that travels at the full radius R. The rest of these pieces of mass, like this one in here, traces out a circle. It definitely traces out a circle, but the circle it traces out is not equal to the radius R. It's got a diminished R value. So how do we determine the angular momentum of an object whose mass is distributed in a way that some of the mass is close to the axis and some of the mass is far away from the axis. That's what we're gonna do in this video. That's the goal and the approach physicists take to this is almost always the same. We say, well, I've got the formula for the angular momentum of a single particle traveling at a single radius, so let's just imagine our continuous object being composed of a bunch of single masses all traveling at a single radius. So if I break this continuous mass up into individual pieces, right? So if I imagine it being broken up into all these little pieces, then if I found the angular momentum of each piece and added it up, I'd get the total angular momentum for the whole object. So let's try this. So the angular momentum of some piece of the object, let's say that little piece of mass, is gonna be, well the mass of that little piece, I'm gonna write m. It's not the entire mass of this entire rod, it'd just be the mass of that small piece times the speed of that piece times the radius that it is at. So to make this clear, let me just write this as like piece one, so this would be m1, v1 and R1 and this would be the angular momentum of that small piece, and you could do this for mass two over here, and you would get that the angular momentum of mass two would be m2, v2 and R2. Now keep in mind that these v's are all gonna be different so the speed out here at the outside edge is gonna be fastest. This speed's not gonna be as great, and this speed closer to the middle is even smaller because they're tracing out smaller circles in the same amount of time as these outside pieces trace out larger circles in the same amount of time. So at this point, you might be worried. You might be like, "This is gonna be really hard. We're gonna have to add all these up. They've all got different speeds. They're all at different radii. How are we gonna do this?" Well, you gotta have faith and something magical is about to happen. So let me show you what happens if we imagine adding all these up. I only draw two. You gotta imagine there's an infinite amount of these so that makes it seem even harder, but imagine breaking this up into an infinite amount of these little discrete masses and considering each individual angular momentum, they'd be very small because this m1 would be an infinitesimal very small piece of mass, and let's add them all up and see what we get. So if we add up all of the mvR's of every piece of mass on this rod, that would be the total angular momentum of the rod. So in other words, this is really just m1, v1, R1 m2, v2, R2 and so on. You'd have an infinite amount of them, right? I can't write them all out 'cause there's an infinite amount. But just imagine that. So what can we possibly do with this? How do we clean this up? When you're doing a physics problem, you don't want to solve an infinite series by writing each term out infinitely. We want a clever way to deal with this, and there's a really clever way to deal with this. Watch this. So if we write this as L equals the sum of mvR. One problem we have is that each mass has a different v. If I can pull things out of this summation, it would help me out 'cause it would simplify things. I could just factor them out, but right now I can't factor out the R, 'cause these all are different radii from the axis. You always measure your radius from the axis here, and they're all at different radii from the axis, and they all have different speeds. But, remember, we like writing quantities in terms of angular variables because the angular variables are the same for every point on this mass. So every point on this rotating rod has a different speed v but they all have the same angular speed omega, so that's a key. That's often what we do and we're gonna do that here. I'm gonna write this as summation of m, but instead of writing v, I'm gonna write this as R times omega. So, remember, for something rotating in the circle, the speed v is gonna be equal to R times omega and that's what I'm gonna substitute down here, so the speed at any point here is the radii of that point times the angular speed of this rod rotating in a circle, and I still have to multiply by the last R here, so this was v. We substituted in what v was, but we have to multiply by R and what do we get? We get that L is gonna be the summation of mR squared omega. And this is great. The omega is the same for every single mass in here. Every single mass travels at the same angular speed, so we could factor this out of the summation. Imagine all these terms would have an omega. We can factor that out, and I could just bring that outside of the summation. So I'll write this as the summation of mR squared, and to make this clear, I'm gonna put parentheses here. It's that summation and then that whole thing times omega, 'cause we're just factoring out omega. And you might not be impressed. You might be like, "All right. Big deal. We've still got an infinite sum in here. What the heck am I gonna do with that?" You don't have to do anything with that. This is where the magic happens. Look at what sum you've got. You've got the sum of all the mR squareds. Remember what mR squared was? Mr squared was the moment of inertia of a point mass, and if I add up all the mR squareds, I get the moment of inertia of the entire mass, this entire object. I get its total moment of inertia. So what we found was a really handy way to write the angular momentum of an object. It's just the moment of inertia of an object, I, times the angular speed of that object. So this is a great formula, and it totally makes sense for this reason. Think about regular momentum, right. Regular momentum, p, was just equal to mv. Well, if you then told me, "Determine the angular momentum," and I didn't wanna go through this derivation, I mighta just been like, "All right, angular momentum, shoot." Well, I'm just gonna replace mass with angular mass, and angular mass, the angular inertia, is just the moment of inertia, and I'll just replace the speed with the angular speed, and look, I just get this formula. So it makes sense because if you replace all the linear quantities with their angular counterpart, you indeed just get the angular momentum of a rotating object. So this is how you do it. If you've got an object, an extended object where the mass is distributed about the entire object, if you just take the moment of inertia of that object and multiply by its angular speed, you get its angular momentum. So, for instance, if this rod has a mass of let's say three kilograms, and that mass is evenly distributed. Let's say the radius of this object is two meters. So that's the distance from the axis to the outside edge. And let's say the angular speed of this object was, let's say 10 radians per second, we can figure out the angular momentum of this rod by saying that the angular momentum is gonna equal the moment of inertia. Well the moment of inertia of a rod about one end is equal to 1/3 mL squared. That's the moment of inertia of a rod about the end, and I then multiply by the angular speed of the object. So if I plug in numbers, I get the angular momentum of this rod is gonna be, I'll use purple here, 1/3 times three kilograms times the length of the object was two meters, and we square that, and then we multiply by the angular speed and that was 10 radians per second, which give us an angular momentum of 40 kilogram meters squared per second. So recapping, if you've got a point mass where all the mass rotates at the same radius and you wanna find the angular momentum, the easiest way to get it is probably with the formula mvR. However, if you have a mass whose mass is distributed throughout the object so that different points on the object are at different radii, the easiest way to get the angular momentum of that object is most likely with the formula I omega, where I is the moment of inertia of the object and omega is the angular velocity of the object.