### Video Transcript

In this video, we’re going to learn
about the conservation of angular momentum. We’ll review what angular momentum
is. We’ll see why it’s conserved. And we’ll learn why the
conservation of angular momentum is important.

To get started, imagine that, as a
hobbyist who enjoys balancing spinning plates, one day as you’re reading the
Guinness Book of World Records, you discover that the world record for the most
spinning plates balanced at one time by one person is six. That doesn’t sound like a lot to
you. And you wonder if you could beat
that record.

With the help of a friend who gets
the plates spinning before you start to balance them, you work your way up to the
record level. But unfortunately, before you can
get there, some of the plates put on earlier start to slow their spin. And eventually they fall. You wonder if there’s anything you
can do regarding the plate rotation rate to help them stay stable for longer. To understand this better, it’s
helpful to know about the conservation of angular momentum.

As a reminder, angular momentum,
which we symbolize with a capital letter 𝐿, is a property of massive systems that
rotate. That means that, any time we have a
massive object that’s rotating about some axis, then that object has angular
momentum. And we’ve seen that that angular
momentum is equal to the cross product of the position vector of the object and its
linear momentum 𝑝. Based on this relationship, we
could say that angular momentum is momentum of rotational motion.

We’ve also seen that there’s
another way to express angular momentum mathematically. Looking back at our diagram, if our
object is rotating with an angular velocity 𝜔 and if the object itself has a moment
of inertia capital 𝐼, then we can say that the object’s angular momentum is equal
to the product of those two variables. Written this way, we’ve expressed
angular momentum in terms of rotational variables only, moment of inertia and
angular velocity.

Speaking of rotational variables,
we can see why angular momentum is conserved by looking at a different relationship
and converting that into rotational variables too. The version of Newton’s second law
of motion that we may be most familiar with says that the net force acting on an
object is equal to the object’s mass times its linear acceleration 𝑎.

If we were to write a rotational
version of the second law, we would replace force with torque 𝜏, we would replace
mass with moment of inertia 𝐼, and we would replace linear acceleration 𝑎 with
angular acceleration 𝛼. And just like in the linear
version, we would be referring to the net torque acting on some object.

Looking at this equation, we
realize we can substitute for the angular acceleration 𝛼, the change in angular
velocity 𝜔 per unit time. That’s because the definition of
acceleration is a change in velocity per unit time. Considering this rewritten
expression for Newton’s second law, the rotational version, we see that the right
side actually looks a lot like the right side of our equation for angular
momentum. Both equations have just the moment
of inertia and the angular velocity 𝜔 in them. But our net torque equation has a
time derivative.

If we assume that our moment of
inertia 𝐼 doesn’t change in time, that means we can factor out the time
derivative. And what’s inside our parentheses
is simply the object’s angular momentum 𝐿. One thing this resulting equation
tells us is that if the net torque on our object is zero, then that means the time
rate of change of angular momentum is also zero. Or, in other words, momentum
doesn’t change in time. It’s conserved.

What we’ve done then is we found
the condition needed in order for angular momentum to be conserved. The requirement is that the net
torque acting on our object is zero. We can write this out by saying
that angular momentum is conserved in a closed system where there’s no net
torque. You might recognize this as similar
to our statement of linear conservation momentum, where we said that linear
conservation is conserved in a closed system with no net force.

One way we can understand
conservation of angular momentum better is to see a few practical examples of this
happening. A classic example of angular
momentum conservation is an ice skater standing in place but rotating quickly. When the skater holds his or her
arms out to full extension, they’ll rotate with a particular speed. We can call it 𝜔 sub 𝑖. But then, if they tuck their arms
in as close as possible to their torso, they’ll rotate with a new speed 𝜔 sub 𝑓,
which is greater than the original angular speed.

We can explain this effect from a
conservation of angular momentum perspective. One way of writing the magnitude of
angular momentum is that it’s equal to an object’s mass times its distance from the
axis of rotation squared times its angular speed. In the case of our ice skater, when
they moved their arms in closer to their chest, that radial distance decreased. And as 𝑟 went down in order for 𝐿
to be conserved, 𝜔 went up. They spun faster. The effect is actually pretty
dramatic, as you might guess from the fact that 𝑟 is squared. So, cutting that value in half will
require 𝜔 to go up by a factor of four in order for angular momentum to stay the
same.

Other examples of angular momentum
conservation are related to objects that move in a stable way because of the angular
momentum they have. For example, if you throw a
frisbee, in order to do so, you give it a spin. Because the frisbee is spinning,
it’s able to fly in a stable path. In a sense, the frisbee tries to
conserve its angular momentum. And the best way to do that is to
keep moving as it has been.

Another example of this is a
bicycle rider who becomes more stable in their motion as they move faster. This is because the spinning wheels
of the bicycle want to conserve their angular momentum as they rotate and,
therefore, resist change to their motion. This is why it’s easier to balance
on a bicycle that’s in motion than one that’s moving very slowly or stopped.

Just like was the case with the
conservation of linear momentum, the conservation of angular momentum is a very
useful concept because it lets us solve problems we might not otherwise be able
to. Let’s see how that works
practically through an example.

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.

Let’s summarize what we’ve learned
about conservation of angular momentum. In this video, we’ve seen that
angular momentum, typically symbolized with a capital 𝐿, is a measure of an
object’s tendency to continue its rotational motion. We’ve also seen that the angular
momentum of a system that experiences no net torques over time stays the same. That is to say, it’s conserved.

And finally, we saw that angular
momentum conservation makes this quantity so useful because it helps us solve
problems, as it did in the case of our spinning ice skater. It’s that problem-solving utility
that makes angular momentum such an important point of focus and so worthwhile to
understand.