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# Change in period and frequency from change in angular velocity: Worked examples

## Video transcript

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 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 ok so the key here is to realize the connection between angular velocity and period and 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 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 this in two different colors we could say T sub large T sub large is going to be equal to 2 pi over it says a large tire spins with angular velocity for Omega so it's going to be 2 pi its angular velocity is 4 Omega so 2 PI over 4 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 2 pi and it's going to have half the angular velocity is a large tire so that half of 4 Omega is 2 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 2 Omega and T sub small the period of the smaller tire that's just going to be hi / Omega and so the red expression right over here is half of this blue expression I could rewrite this as being equal to 1/2 times 1/2 times pi / Omega or another way of writing this I could write T sub large T sub large is equal to 1/2 times this expression right over here is a 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 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 2 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 have to complete two pi radians to complete a cycle and then we could divide that by how quickly we are how quickly our angle is increasing and so there you have it that's our period is 2pi 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 2 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 frequency let's call it frequency initial initial is going to be equal to her angular velocity is 2 Omega it's going to be 2 Omega over 2 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 phenomenon in the in other videos is going to be so decreasing her angular velocity 2 Omega so now her angular velocity is omega and it's going to be over 2 pi so how do these to come out of these compare well if I write her initial frequency I could rewrite it as initial frequency initial frequency is equal to 2 times Omega let me do that another color 2 times Omega over 2 pi write 2 times all of this business right over here which is the exact same thing which is equal to 2 times our final frequency 2 times our frequency final and or another way of thinking about it her frequency final her frequency final if I divide this and this by 2 is going to be equal to 1/2 of her frequency initial is going to be equal to 1/2 of her initial frequency frequency initial if your initial frequency is twice your final then your final is going to be 1/2 your initial I could just divide both sides by 2 so once again how does the frequency of her spin change well it looks like her frequency goes by 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 is decreasing 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 1/2
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