 Now we've seen the equation for angular momentum and torque and was explicit about the point that you're going to get different equations depending on what your reference point is. Now, let's have a look at this problem here. Imagine this is a rod and from it there's another rod that is angled and this sits on some form of a mechanism so that this can turn around and around and around. So this arm is rigid, this arm is rigid, there's a point mass there and it just spins around at angles like this and so this just spins around. We want to know point A for this point A, what is its angular momentum as it goes around and what is the torque on the system? Now I have a choice as to what my point of reference is going to be. Let's choose very cleverly first In a clever way, let's choose this point right here as our point O. That is the point with which we're going to take angular momentum and torque with with respect to that point. Let's do that. If we do that, we've got to consider this as our world reference frame and this is our intermediate frame. So let's construct then back to square one where we start at this course. We've got to have a reference frame there, an intermediate frame. If this is our world frame and we make this the position vector. Now I forgot to say there's an angle between this right angle here and the rod, so it's pointed up at this angle theta. The length of this section here is L. That makes this L sine theta and this L cosine theta. It's not important for us right now. We're just going to construct the fact that we just have a position vector in this direction of our hat. That would be an our hat for us which means into the board if we think it spins into the board and out of the board that we have theta hat. That we have theta hat into the board there. So it points straight in and our hat is on this side. If we now with respect to this point write R vector A, our point A with respect to O. What would that be? What would that be at the moment? Well as we've drawn it here it'll be this length R which is L cosine of theta from that point to that point R hat. That's what we have. And that's what we have. Now remember that angular momentum if we're just going to take angular momentum of A with respect to O that is going to be R of A with respect to O times P of A with respect to O. Cross product of those two vectors. What is this? This is M times the second derivative of A with respect to O. But that's not true. That's first derivative. Linear momentum. First derivative of A with respect to O. M times V. That's all we have there. So we will need at least the first derivative there. And if we think of the torque of A with respect to O, the sum of the torques there is going to be R of A with respect to O cross whatever forces they are at A with respect to O. And what is force? Well that is going to be mass times the second derivative of the position vector. So we're going to need its first derivative and its second derivative. That was one way. We could also have said it is the first time derivative of angular momentum plus that funny term V cross P. But Sydney, this would be one way and we see we have R. We need its first derivative and its second derivative. If you're dealing with a problem like this little shortcut, let's get those out of the way right now. So let's get R A with respect to O dot. That's first time derivative. So what are we going to get? We're going to get R dot R hat plus we've done this a million times the d dt of R hat is going to be theta dot whatever the angular velocity is in that direction. So theta dot theta hat. And that's it. Now, if you think about it, if this is a rigid messless rod, this R, the length of this vector never changes. R is a constant which makes this term zero. So if we're dealing with problems in such as this, we can immediately take away you can immediately take away this term. This term is of no consequence to us. If R is not, if R is not constant, this R is not constant. You cannot take away that. Let's get R double dot of A with respect to O. Now, I don't need this term. This is a problem where we are going to say that R equals a constant, make a constant one. In other words, we only have to get the first derivative of that. That's simple. It's R dot theta dot theta hat plus we're going to add R theta double dot theta hat minus R theta dot squared R hat. Remember the DDT of theta dot? It's negative R theta dot R hat. Now, we've already got a theta dot there, so it becomes theta dot squared. And again, I see there is a term there which I'm not interested in because it is zero. If this is a constant, its first derivative of a constant is zero. So I've got those two. All I now need to do is to do this. I'm going to say the angular momentum of point A with respect to O is going to be R R hat cross P. This is M times V. This is M times this. So it's M R theta dot theta hat. Simple as that. And this leaves us with R A with respect to O being, well, I have an M, I have an R squared theta dot, and what is R hat cross theta hat? Well, that's K hat. That is as simple as that. What about the angular momentum? Well, I can do it in two ways now. I can just take the first derivative of this. M is a constant. R is a constant. Theta dot is not a constant. So let's do it that way. DDT. Let's do the DDT of H bar A with respect to O. Remember, that gives us, that is the torque, some of the torques around of point A with respect to point O. And that is actually remember plus V cross P, but V and P is parallel to each other, the velocity, the linear velocity and the linear momentum, obviously, so that term would fall away. So all I have is this. Now, taking the first derivative of this, remembering M is a constant. I can bring that out of the derivative. R squared is a constant. K hat is a constant. So it's just the DDT of theta dot. And what is that? That leaves me with M R squared theta double dot K hat. As simple as that. So now I have the angular momentum and the torque for this with respect to this point here. Let's make my life very, very easy. You'll also see if you take R cross F. So if I were to take this whole term here, this bit here, and I crossed it with, multiplied with M. So it's this cross M multiplied by this. I was going to end up with exactly the same term. No issues there. So multiple ways to do this. If you do these two a little trick, if you know you're going to use both of them, do them before, throw away these terms and makes this whole problem a lot easier. And the next lecture I want to show you, if we move our world frame to down here, we take the angular momentum of A with respect to O down here and the torque, that becomes a lot more complicated and tells us, actually tells us a bit more about the system.