Change in period and frequency from change in angular velocity: Worked examples | Khan Academy

We're told that a large tire spins with angular velocity (4 \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_{\text{large}}) of the large tire compare with the period (T_{\text{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.

Ok, so the key here is to realize the connection between angular velocity and period. Instead of just blindly memorizing a 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'm going in a circle around like this—in order to do one complete lap around the circle or a complete one cycle, I have to cover (2\pi) radians. So (2\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.

So we can say (T_{\text{large}}) (I'll do this in two different colors). We could say (T_{\text{large}} = \frac{2\pi}{\text{angular velocity of the large tire}}). A large tire spins with angular velocity (4 \Omega), so it's going to be (T_{\text{large}} = \frac{2\pi}{4 \Omega}).

Then for (T_{\text{small}}), a smaller tire spins with half the angular velocity, so (T_{\text{small}} = \frac{2\pi}{\text{angular velocity of the small tire}}). Half of (4 \Omega) is (2 \Omega), so (T_{\text{small}} = \frac{2\pi}{2 \Omega}).

How do these two things compare? Well, it might be helpful to just simplify these expressions a little bit. So (T_{\text{large}}) (the period of the large tire) is going to be (\frac{\pi}{2 \Omega}) and (T_{\text{small}}) (the period of the smaller tire) is just going to be (\frac{\pi}{\Omega}).

So the red expression right over here is half of this blue expression. I could rewrite this as being equal to (\frac{1}{2} \times T_{\text{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 (\frac{1}{2}) the period of the smaller tire.

Now it's always nice, if you have the time—if you know if you're on time pressure—to just think about whether that makes sense. So a large tire spins with an angular velocity of (4 \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 small 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 (2 \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 the period from the last question is given by (T = \frac{2\pi}{\text{angular velocity}}). If we want frequency, frequency is just the reciprocal of the period. So frequency is just going to be (\frac{\Omega}{2\pi}). This is how many cycles we can complete in a second.

At first, the ice skater spins with an angular velocity of (2 \Omega). So let's say (f_{\text{initial}} = \frac{2\Omega}{2\pi}).

Then her frequency final, after she puts her arms away from her body (decreasing her angular velocity), will be given by (f_{\text{final}}=\frac{\Omega}{2\pi}).

How do these two compare? Well, if I write her initial frequency, I could rewrite it as:
(f_{\text{initial}} = \frac{2 \times 2\Omega}{2\pi}), which is the same as (2 \times f_{\text{final}}).

Another way of thinking about it is that her frequency final, if I divide both sides by (2), is going to be equal to (\frac{1}{2}) of her initial frequency. If your initial frequency is twice your final, then your final is going to be (\frac{1}{2}) your initial.

So, how does the frequency of her spin change? Well, it looks like her frequency goes down by (\frac{1}{2}), and that makes sense. If your angular velocity is going down by half, you're 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 by a factor of two. It is having decreasing by a factor of two is the same thing as saying your frequency gets multiplied by (\frac{1}{2}).