Angular Momentum of a Point Mass

Concept: Angular Momentum of a Point Mass

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Video Transcript

Hey guys. So in this video I want to talk about the angular momentum L of a point mass. Now remember that in rotational problems, we can have either point mass which is a tiny object of no shape of no radius, no volume or you can have a rigid body, an extended body which is an object of a known shape that has a radius, has a volume and depending on whether you have a point mass or a rigid body questions are gonna be a little bit different. So let's look at the angular momentum of a point mass, let's check it out. So first I want to, I want to talk about a point mass in a circular path, a point mass tiny little m in a circular path like this about a central axis right there, it's going to have a rotational speed Omega and it has a linear equivalence v10, v10 the tangential velocity that you have when you're spinning. Now this does not mean, this does not mean that you have two types of motions you only have one type of motion which is rotational motion, okay? Just because you have a W and a V it doesn't mean you have linear and angular motion you only have rotational motion, that's why I call this a linear equivalence, okay? So that's a first distinction I want to make, I also want to point out that linear momentum and rotational momentum are not going to give you the same number, are not going to give you the same number. So because I have an Omega I can find a rotational momentum, rotational momentum l equals I Omega. So if I have Omega, I have this, well, if I'm going in a circle I also have a equivalent velocity V 10. So because I have a V that's equivalent to the W, I can find a linear momentum as well. So if I have a W, I can calculate L if I have a V, I can calculate p but these two will not give you the same number and that's because linear momentum remember is absolute but angular momentum is relative, it depends on the axis. So p is always going to be the same if your V is the same but if you change the radius of your rotation your L will change. So these numbers will not be equivalent to each other, they'll not be the same number, okay? If you calculate linear and angular momenta, momenta is just plural for momentum, linear momentum is p, angular momentum is L, for a point mass you're going to get different equations and different numbers, we need a quick example here so I can show you what I mean by that and you see how this stuff works. So I have a small two kilogram object so mass equals 2 small tells me that this is going to be a point mass, it's not going to have a shape, it spins horizontally around the vertical axis. So spinning horizontally means you spin like this around the vertical axis is just, imagine that you have an imaginary line that you spin around that line and that line is vertical, okay? So I'm going to draw the axis here and you're sort of spinning like this, okay? So here's the object m right there, you do this at a rate of 3 radians per second, 3 radians per second it's going to be your Omega, you know that because of the units radians per second so Omega equals 3 and it says maintaining a constant distance of 4 to the axis. So it mean that your distance to the axis is going to be 4, distance to the axis is little r. So I'm going to write that little r is 4. Alright, so we want to know what is the object's linear momentum and what is the object's angular momentum about its central axis. So for A we're looking for P, P equals mv and then for B we're looking for l, l equals I Omega, okay? Now for the purposes of solving this question before I find the numbers I actually want to change up these equations, I can change L to make it look kind of like p and I want to do this to show that even though L can look like p it's not going to be exactly the same, okay? You see what I mean once I get there. So what I'm going to do here is I'm going to replace I with the moment of inertia equation, remember moment of inertia of a point mass is given by mr squared where r is distance to the center to the axis of rotation so I'm gonna write this as mr squared. Now Omega is related to V, right? So if you're spinning like this and you'll have an Omega and you have a V they are related by, there's an r, r vector is the length, the distance, right? They're related by r equals r Omega. So if I rewrite Omega, Omega is going to be v over r, okay? So instead of Omega I'm going to have v over r, again I'm doing this to show you something, look what happens here, r cancels with r and then I have mvr, I'm going to change it around, I'm going to write instead of mvr and the point that I will make here is that p is mv, p is mv and L is mvr therefore not the same equation and not the same number, okay? So now we can plug stuff in, mass is 2, the velocity, the velocity would have to calculate that. So v equals rw. So r is distance 4, w is 3, so v is 12, so m is 2 and then this is 12, so this is going to be 24 and it's kilograms times meters per second. Now for L I have mass which is 2, I have 12 and the velocity in r is the distance which is 4. So this is going to be 96 kilograms times meters squared over seconds, okay? So notice how these are different equations and they are different answers, okay? The only case in which there will be the same is if the r happens to be 1 which is just a gigantic coincidence and then in that case there will happen to be the same but there's nothing special about an r equals 1 other than you get the same numerical value, okay? So that's it for this one, the last point, I'm going to make here is that this equation that we've got here, that L. So L is I Omega but you can also have L as mvr for a point mass. So this equation L, mvr is also used for angular momentum L of an object in linear motion about an axis of rotation. So if you ever get asked the question that says find the, find the angular momentum of an object about an axis of rotation, this is how you would do it. Okay we're going to do more of this soon but I want to quickly show you an example, let's say this is one of the classic ones, let's say you have this, a bar being supported by a fulcrum here and then a little object is falling here, right? So the object is about to hit. So when the object is about to hit it has a velocity of let's say of 10, okay? So even though this object is moving in linear motion you can ask what is the angular momentum of this object relative to the axis of rotation, the axis of rotation is here because this is a point about which this could spin, you would write that L is mvr, you will be given the the m, v is 10 and r would be this distance here, okay? So this equation here can be used if you want to find the angular momentum of an object that has linear motion, okay? So that's a finished one, hope this makes sense, let me know give any questions and let's get going.

Problem: The Earth has mass 5.97 x 1024 kg, radius 6.37 x 106 m. The Earth-Sun distance is 1.5 x 1011 m. Calculate its angular momentum as it spins around itself. Treat the Earth as a solid sphere of uniform mass distribution. 

BONUS 1: Treating the Earth as a point mass, calculate its angular momentum as it spins around the Sun. 

BONUS 2: Does the Earth have linear momentum as it spins around (i) itself; (ii) the Sun?

8m

Problem: A system is made of two small, 3 kg masses attached to the ends of a 5 kg, 4 m long, thin rod. The system rotates with 180 RPM about an axis perpendicular to the rod and through one of its ends, as shown. Calculate the system’s angular momentum about its axis.

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