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Current time:0:00Total duration:6:48

- [Instructor] We're told
that a large tire spins with angular velocity four omega. A smaller tire spins with
half the angular velocity. I'm assuming half the angular velocity of the large tire. How does the period T
sub-large of the large tire compare with the period T
sub-small of the small tire? So pause this video and see if you can figure that out and figure out which choice you would pick. Okay, so the key here is
to realize the connection between angular velocity and period. And instead of just blindly
memorizing the formula, I always like to reason
it through a little bit. We know that period is equal to, well think about it. In order to complete
one cycle, if I'm doing uniform circular motion, if I am going in a circle around like this, in order to do one complete
lap around the circle, or to complete one cycle, I
have to cover two pi radians. So two pi radians is what I need to cover, and then I divide that
by my angular velocity, how fast am I going through the radians. So that's how I like to
reason through this formula that connects angular
velocity, or the magnitude of angular velocity, and the period. And so, we can say T sub-large, I'll do some two different colors, we can say T sub-large, T sub-large is going to
be equal to two pi over, it says a large tire spins with angular velocity four omega, so it's going to be two pi, its angular velocity is four omega, so two pi over four omega. And then T sub-small, a smaller tire spins with half the angular
velocity, so T sub-small, T sub-small is going
to be equal to two pi, and it's going to have
half the angular velocity as the large tire, so
that half of four omega is two omega. So how do these two things compare? Well it might be helpful to just simplify these expressions a little bit. So T sub-large, the
period of the large tire, is going to be pi over two omega, and T sub-small, the
period of the smaller tire, that's just going to be pi over omega. And so, the red expression right over here is half of this blue expression. I can rewrite this as being
equal to one half times, one half times pi over omega. Or another way of writing this, I could write T sub-large, T sub-large is equal to one half times, this expression right over here is the period of the smaller
tire, so T sub-small. Now which of these
choices match up to that? Well it is this one right over here: the period of the larger
tire is going to be one half the period of the smaller tire. Now, it's always nice
if you have the time, if you aren't under time pressure, to just think about
whether that makes sense. So a large tire spins
with an angular velocity of four omega; the smaller tire spins with half the angular velocity. So, if it has half the angular velocity, it's rotating half as fast. If it's rotating half as fast, it would take twice as
long to complete one cycle. So the smaller tire is
gonna take twice as long, or you could view it as the large tire takes half as long as the small tire. So that makes sense. Let's do another example. An ice skater spins with
angular velocity two omega. She brings her arms away from her body, decreasing her angular velocity to omega. How does the frequency of her spin change? Once again pause this video and see if you can figure that out on your own. Well let's just think about how frequency is connected to angular velocity. We already know that period
from the last question is equal to, or you'll have
to complete two pi radians to complete a cycle, and
then we can divide that by how quickly our angle is increasing. And so, there you have it. That's our period. It's two pi over our angular velocity. And if we want frequency, frequency is just the reciprocal of period. So frequency is just going
to be omega over two pi. This is how many cycles we
can complete in a second. And so, at first the ice skater spins with an angular velocity of two omega. So let's say frequency, let's call it frequency initial, initial, is going to be equal to, her angular velocity is two omega, it's going to be two omega over two pi. And then her frequency final I'll say, frequency final, so after she puts her
arms away from her body decreasing her angular velocity, and we'll talk more about that phenomena in other videos, is going to be, so decreasing her angular
velocity to omega, so now her angular velocity is omega, and it's gonna be over two pi. So do these compare? Well if I write her initial frequency, I can rewrite it as initial frequency, initial frequency, is
equal to two times omega, I'm gonna do that in another color, two times omega over two pi. Right? Two times all of this
business right over here. Which is the exact same thing, which is equal to two
times our final frequency. Two times our frequency final. Or another way of thinking about it, her frequency final, her frequency final, if I divide this and this by two, is going to be equal to one half of her frequency initial, is going to be equal to one half of her initial frequency,
frequency initial. If your initial frequency
is twice your final, then your final is going to
be one half your initial. I could just divide both sides by two. So once again how does the
frequency of her spin change? Well it looks like her frequency goes by half, and that makes sense. If your angular velocity
is going down by half, you are rotating half as fast and so you're going to be able to complete half as many cycles per second. So it makes sense that we are decreasing, our frequency is decreasing by a factor of two, it is halving. Decreasing by a factor
of two is the same thing as saying your frequency
gets multiplied by one half.

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