9.1 Uniform Circular Motion

9.1 Uniform Circular Motion


When we analyzed how the
position vector changed, we know that the velocity
for circular motion is given by the
radius times the rate that the angle is changing. And it points tangential
to the circle. So let’s draw a few
characteristic arrows to show that. At this point, we’ll draw
these pictures with d theta dt positive. So the velocity
points like that. It points like this. It points like that. And these are all the velocity
vectors at different times. Notice that if we make–
consider the special case in which d theta dt is a
constant, in that instance, the magnitude of
the velocity, v, is given by r magnitude
of d theta dt. And that is also a constant. But the velocity vector
is changing direction. And we know by definition
that the acceleration is the derivative of velocity. And so what we see
here is where we have a vector that’s constant
in magnitude but changing direction. And we now want to
calculate the derivative in this special case. We refer to this case as
uniform circular motion. So this special case is often
called uniform circular motion. OK. How do we calculate the
derivative of the velocity? Well, recall that the
velocity vector, r d theta dt– those are all constants–
because it’s in the theta hat direction, once again,
will decompose theta hat into its Cartesian components. You see it has a minus i hat
component and a plus j hat component. The i hat component
is opposite the angle. So we have minus sine theta
of t i hat plus cosine theta of t j hat. So when I differentiate
the velocity in time, this piece is constant, so I’m
only again applying the chain rule to these two functions. So I have r, d theta dt. And I differentiate sine. I get cosine with a minus sign. So I have minus cosine theta. I’ll keep the function of
t, just so that you can see that– d theta dt i hat. Over here, the
derivative of cosine is minus sine d theta dt. That’s the chain rule– sign
of theta dt, d theta dt, j hat. And now I have this
common d theta dt term, and I can pull it out. And I’ll square it. Now whether do you think that
dt is positive or negative, the square is always
positive, so this quantity is always positive. And inside I have–
I’m also going to pull the minus sign out. And I have cosine theta of t i
hat plus sine theta of t j hat. Now what we have here is
the unit vector r hat t. r hat has a cosine adjacent in
the i hat direction and a sine component in the
H Hut direction. So our acceleration–

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