### Video Transcript

An ice skater is preparing for a
jump in which he will rotate while in the air. When he is on the ground with his
arms extended, his moment of inertia is 2.2 kilograms meter squared, and he is
spinning at 0.31 revolutions per second. He launches himself into the air at
a speed of 12.6 meters per second at an angle of 45 degrees above the
horizontal. At the moment that he leaves the
ground, he contracts his arms, taking negligible time, and changes his moment of
inertia to 0.62 kilograms meter squared. How many revolutions can he
complete while airborne?

We’ll call this number of
revolutions capital 𝑁. And we’ll start by drawing a sketch
of the situation. When the ice skater jumps, he has
an initial moment of inertia we’ve called 𝐼 sub 𝑖 and an initial angular speed
we’ve called 𝜔 sub 𝑖, both given to us in the problem statement. He launches himself at an angle
we’ve called 𝜃 above the horizontal, with an initial take-off speed we’ve called
𝑣.

Just as he launches off the ice, he
pulls his arms into his chest and achieves a final moment of inertia we’ve called 𝐼
sub 𝑓, also given to us. While the skater is in the air,
he’s rotating at an angular speed we’ve called 𝜔 sub 𝑓 and, after rotating at this
speed while he’s airborne, lands back on the ice. Over that time and spinning at that
rate, he completes some number of revolutions we’ve called capital 𝑁. And that’s what we wanna solve
for.

We’ll begin by solving for 𝜔 sub
𝑓, the skater’s angular speed while he’s in the air. And to do that, we’ll rely on the
principle of the conservation of angular momentum. In a closed system that experiences
no net torques, we can say that the system’s angular momentum is equal at any one
time to the angular momentum of the system at any other time. That is, the system’s initial
angular momentum, wherever we define initial to be, is equal to its final angular
momentum.

In our case, we define the initial
moment to be a time before the skater has left the ice and the final time to be
after the skater has left the ice and changed his moment of inertia. So, we can write that 𝐿 sub 𝑖 is
equal to 𝐿 sub 𝑓. And recalling that the magnitude of
angular momentum is equal to moment of inertia multiplied by an object’s angular
speed, we can expand this statement to say that 𝐼 sub 𝑖 times 𝜔 sub 𝑖 is equal
to 𝐼 sub 𝑓 times 𝜔 sub 𝑓.

In all this, it’s 𝜔 sub 𝑓 we want
to solve for. We see it’s equal to the ratio of
the initial moment of inertia to the final moment of inertia multiplied by the
initial angular speed. We’re given all these values in the
problem statement, so we can plug in and solve for 𝜔 sub 𝑓. Plugging these values in and
entering them on our calculator, we find 𝜔 sub 𝑓 is 1.1 revolutions per
second. That’s the angular speed of the
skater when he’s in the air.

Knowing that, if we knew how long
the skater was in the air, then we could figure out how many revolutions he
performed and solve for capital 𝑁. While the skater is airborne, he’s
essentially a projectile, under the rules of projectile motion. If we call the total horizontal
range that the skater travels capital 𝑅, there’s a relationship we can recall which
helps us solve for 𝑅 based on the initial launch speed 𝑣 and the angle 𝜃.

Sometimes called the range
equation, this relationship says that the horizontal range of an object which is
launched and lands at the same elevation is equal to the square of its initial
speed, 𝑣 sub zero, times the sin of two times its launch angle 𝜃, all divided by
the acceleration due to gravity 𝑔. If we let 𝑔 be exactly 9.8 meters
per second squared, then we can say that the range our ice skater travels while in
the air is equal to 𝑣 squared sin two 𝜃 over 𝑔, or 12.6 meters per second squared
times the sin of two times 45 degrees, or the sin of 90 degrees, all divided by 9.8
meters per second squared.

We can simplify this last equation
a bit because we know that the sin of 90 degrees is equal to one. Before we calculate this result
though, let’s consider how it fits into the bigger picture of solving for 𝑁. We can recall that the average
speed of an object is equal to the distance it travels divided by the time it takes
to travel that distance. This range value 𝑅 that we’re
calculating is the distance the skater travels in the horizontal direction.

That means that if we were to apply
this relationship to our scenario, we would use the ice skater’s horizontal speed 𝑣
times the cos of 𝜃 and set that equal to the horizontal distance traveled 𝑅
divided by the time the skater is in the air 𝑡. Really, it’s not 𝑅 we wanna solve
for, but 𝑡, because if we have 𝑡 and we multiply it by 𝜔 sub 𝑓, we’ll get a
total number of revolutions the skater has traveled. In other words, we’ll have solved
for capital 𝑁.

We can rearrange and say that 𝑡 is
equal to 𝑅 divided by 𝑣 times the cos of 𝜃. And now rather than calculating the
range 𝑅, we can calculate the time 𝑡 by dividing the range 𝑅 expression by 𝑣
times the cos of 𝜃. So, now we’re solving for 𝑡, which
is equal to 𝑣 squared over 𝑔 times 𝑣 times the cos of 45 degrees. And we see that a factor of the
speed 𝑣 cancels out from numerator and denominator. Entering this simplified fraction
on our calculator, we find that 𝑡 is approximately 1.818 seconds. That’s the amount of time that the
skater is in the air.

We’re now ready to solve for
capital 𝑁, which is 𝑡 times 𝜔𝑓, or 1.818 seconds times 1.1 revolutions per
second. To two significant figures, this is
equal to 2.0 revolutions. That’s how many turns the skater
would make while airborne.