Now that we've gotten familiar with this concept of rotations and displacements more generally, it's time to start thinking about the rate of change of rotations. And that leads us to the concept of an angular velocity vector. What does it mean to differentiate a rotation and get a velocity? We know the analogue for position vectors, you differentiate a position vector, you get a velocity. Now we want to take a three by three rotation matrix and differentiate that. Well, if you take this matrix and you differentiate it, the first thing you want to recognize is that this matrix is not just a bunch of numbers it's actually Orthogon and there are two relationships which govern the Orthogonality. Let's differentiate both sides of each equation, if you do that you'll essentially use the product rule. And you come up with two identities that relate to the derivative of the rotation matrix and the transpose of a rotation matrix. And this will tell you that R transpose R dot, where R dot is a derivative of R, and R dot R transpose are both skewed symmetric. In other words, there's only three independent elements in these q symmetric matrices. So rather than think in terms of the derivative of a rotation matrix, we will think in terms of the derivative, pre-multiplied by R transpose, or post-multiplied by R transpose. Back to our canonical example for a rigid body with a generic position vector and its rotated position, which we call P prime. Once again, we'll make sure that the origins are the same, and we'll consider the displacement from P to P prime. Recall that we use the position vector P to denote the coordinates of the point P, with respect to A1, A2, A3, and the vector q to denote the coordinates of the point p prime with respect to A1, A2, A3. I want us to remember now that the vector P is a three by one vector that consists of coordinates of P in a body fixed frame. Now if the rigid body is rigid and stays rigid, this vector P does not change. The only thing that changes as a rigid body rotates is a vector q. As the rigid body rotates, the rotation matrix changes as a function of time, and the vector q changes as a function of time. But the vector p stays constant. So let's take this equation and differentiate this. If you differentiate both sides and, again, take advantage of the fact that p remains constant, you will see that the derivatives that are appear are only the derivatives of q and the derivatives of R. [NOISE] So p as we've seen before is the position in a body fixed frame. Q dot on the other hand is the velocity in an inertial frame. If I pre multiply both sides by the transpose of the rotation matrix I get a familiar form on the right hand side. On the left hand side, it's the velocity and the body fix frame, Q dot. The velocity in the inertia frame is being transformed back to the body fix frame. And the right hand side, the familiar quantity that I see is R transpose R dot. And that encodes the angular velocity in a body-fixed frame. Let's call that omega hat, and attach a superscript b to denote that it's a body-fixed angular velocity. Again, it is the angular velocity of the rigid body, but we've chosen to write the components in a body-fixed frame, which is not shown here. Another way of writing the equation relating the velocity in inertial frame and the position in the body-fixed frame is by rewriting p on the right-hand side in terms of R transposing q. So now let's take a look at the quantities on both sides of this equation. Q dot is the velocity in inertial frame. We've now transformed it back to the body-fixed frame. But on the righ thand side you see another familiar quantity R dot R transpose which is a skew symmetric matrix. This encodes the angular velocity in inertial frame. We call that omega hat with a superscript s to denote the fact that it is not in the body fix frame but instead it's a spacial angular velocity. We've seen before that skewed symmetric matrices encode cross products. So what you're seeing in these two equations is essentially our ability to generate velocities by taking the cross product of an angular velocity vector with a position vector. In the first equation, it's the angular velocity in the body-fixed frame which yields the velocity in the body-fixed frame. In the second equation, it's the angular velocity in the inertial frame which then yields the velocity in the inertial frame. So we have two different representations of the same angular velocity vector. The first one is written in terms of basis vectors on the body-fixed frame, and the second one is written in terms of basis vectors in inertial frame. [SOUND] So here's an exercise worth doing. Let's consider a simple rotation, a rotation about the z axis. This is easy to visualize, and it's also easy to write down. Let's take the transpose of this. And then let's differentiate R with the respect to Time. So R dot will only depend on the derivative of theta. So as this angle changes R changes. And you can write R dot as a function of theta dot by simply pre multiplying it by a matrix that has mostly zeroes except for cosine theta and sine theta. So if you do the same computations for this very simple matrix and derive the expressions for R transpose R dot, and R dot R transpose you'll discover those two matrices are the same. And this happens in this very special case where axis of rotation is constant. In this particular case this symmetric matrix corresponds to the 0,0,1 vector which is the z axis. And this is something you should have expected. If you rotate a rigid body about the z axis and the axis of rotation is constant, then clearly, the angular velocity vector will also be along the z axis. What happens if your rotation is obtained by composing two rotations? In this case the rotations are about the z axis through theta. About the x axis through phi. Now if I do the computations I differentiate r. I pre multiply by r transpose to get the body fixed angle of velocity I'll get two terms. One that depends only on the rate of change of Rz and the second that depends only on the rate of change of Rx. The same thing is true if I use the spatial angular velocity which is R dot R transpose I get two terms. One that depends on the rate of change for Rz. And the second that depends on the rate of change of Rx. You can see that these two expressions are different. In both case they consist of two terms. One depending on theta dot and the second depending on phi dot but because the axis of rotation is not fixed the two expressions are different. The body fixed angular velocity and the spatial angular velocity have different expressions.
