- [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.