The time-dependent position of a point object which moves coun...

Question

The time-dependent position of a point object which moves counterclockwise along the circumference of a circle (radius R) in the xy plane with constant speed υ is given by $\vec{r}$ = î R cos ωt + ĵ R sin ωt where the constant ω = v/R. Determine the velocity $\vec{v}$ and angular velocity $\vec{w}$ of this object and then show that these three vectors obey the relation $\vec{v} = \vec{ω} \times \vec{r}$ .

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with the relationship between velocity and angular velocity. Here, we have a satellite in a stable circular counterclockwise orbit around the earth moving at a constant speed. The satellite's position is given by um this equation here where the position R is equal to capital R cosine Omega T I hat plus um R Sine Omega TJ hat where R is the radius of the orbit and Omega is equal to the speed divided by the radius the V over R and that is the angle of velocity with respect to the earth assuming the orbit is counterclockwise um determine the velocity V, the angular velocity omega of the satellite and is V equal to Omega cross R. We're given force choices here. Um Choice A is the velocity is equal to minus V sine Omega T I hat plus V cosine Omega TJ hat. The angular velocity is equal to V over RK hat and yes choice B the velocity is equal to V Sine Omega T I hat minus V cosine Omega TJ hat. Um The angular velocity is equal to um V over RK hat and Yes. C the velocity is equal to V sine Omega T I hat minus V cosine Omega TJ hat. The angular velocity is equal to V over R I hat. No. And D the velocity is equal to V sine Omega T I hat plus V cosine Omega TJ hat. The angular velocity is equal to V over RJ hat. And yes, so we need to find um three things in this um problem, the velocity of the velocity and we need to calculate um omega cross R to see if it's equal to V. Let's start with the velocity. So if you recall the velocity is just the time derivative of the position equation, so it's gonna be just the time derivative of the position. So we'll have DDT of our position equation. So we have the capital R cosine Omega T I had plus R find Omega T J hat. So this is a simple enough derivative. So as we put, move the derivative through here, um R is constant. So um it just moves through the derivative. The derivative cosine is negative sine of negative sign omega T. And we need to take the um derivative of the argument of that cosine function derivative of Omega T is just omega that goes outside the trick function um I hat is a constant vector. So it moves through the derivative have I had there, then we can move on to the next term um are still a constant So it moves through the derivative, the derivative of sine is cosine, we'll have cosine omega T. Again, the derivative of that argument of mega T is just omega and we have J hat moving through derivative because it's also a constant um vector. And after doing that, um you notice that we have this Omega R in both terms. And if we look up here at our definition, um Omega is equal to V over R, that means that Omega times R is equal to V. So we can actually rewrite this as minus V thine Omega T I hat LV cosine Omega T they had. And so that is our expression for the velocity, let's move on to the angular velocity. And we're looking for the vector for the angular velocity. Now to get the angular velocity, we need to actually use the right hand rule. And we're told that this orbit is moving counterclockwise. So to use the right hand rule, we need to curve our fingers in the direction of the orbit. So curve your fingers in a counterclockwise direction and your thumb is going to point in the direction of the angular velocity. In this case, your thumb should be pointing towards you or out of the page and that is going to be the positive K hat direction. So our omega here is just gonna be whatever its magnitude is, which we just label as Omega and it's gonna be going in the K hat direction. But just like uh we mentioned before that um Omega is equal to V over R, we can rewrite this as V over R. Hey hat. And so that is our angular velocity. The last part of this question, what's this to find out if our velocity is equal to our angular velocity crossed R? So we'll just need to calculate the omega cross R and see if we come up with our expression for the velocity. So our Omega cross R. So for Omega, I'm going to have RV over R in the khat direction and went across that with our expression four R, that's gonna be R cosine Omega T I had plus R sign um Omega E, they had, you could think of a cross product as multiplication. So I can actually distribute um this to both terms inside those brackets. And when I do that, um I get V over R times R which gives me just B I have the cosine Omega T and then I'll have a hat, Ross, a hat. We'll do the same thing for the second term. I'll have V over R times R. So again, I'll get just V and I'll have sign Omega T and here I'll have the K hat. Cross J hat. It's important to keep the order of that cross product, correct. So I had a K hat on the left and so I should have a K hat on the left side of both of those cross terms. Now, I need to just perform those cross products for those unit vectors. And um in case you need a little help remembering um what these cross products actually equal. Um You can actually just write out IJK IJK and then write out IJ again. And so I cross J is equal to K J cross K is equal to IK cross, I is equal to J. And so that'll give you all the permutations possible. So if I look here, I have K cross, I, well, that is going to give me J hat. This first term is just gonna be V cosine Omega T Jay hat. And the second term here, I have K cross J. If I look at my little helper here, I have J cross K gives me I hat. But remember if you flip um the two terms in a cross product, you pick up a negative sign. So K cross J is equal to negative I hat. The second term is gonna come in with a negative sign. So I have negative Bs Omega T I had. And if I look, that's exactly what we had or are velocity vector, but this year does indeed equal our velocity vector. So if I come back up here, my choices and I look at each of those choices, answer a as the correct choice for all three. So that's the um answer to this question. So I hope that this has been helpful and hopefully jog some of your memory um about cross products and um dealing with vectors. So I hope this has helped and I'll see you in the next video.

Key Concepts

Circular Motion

Velocity and Angular Velocity

Cross Product of Vectors

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