### Video Transcript

The wheel of a car initially has an
angular momentum of 12 kilograms meter squared per second. The car accelerates, and a constant
torque of 1.8 newton meters is applied to the wheel for 15 seconds. What is the angular momentum of the
wheel after the car’s acceleration?

Let’s say that this is our car’s
wheel. And we know that, initially, it’s
rotating such that the wheel has an angular momentum of 12 kilograms meter squared
per second. We’ll call that value 𝐿 sub i. We’re then told that a constant
torque is applied to the wheel. Now, at first, we may wonder, is
that torque applied in the direction the wheel is already rotating or is it applied
in the opposite way?

Our problem statement answers this
question by telling us that the car accelerates as this torque is being applied. This means the applied torque was
indeed in the same direction as that in which the wheel was originally spinning. We can call this torque magnitude
of 1.8 newton meters 𝜏. And we know it was applied for a
time interval of 15 seconds. We’ll call that Δ𝑡.

Knowing all this, we want to know
what is the angular momentum of the car’s wheel after acceleration. We can call this value 𝐿 sub
f. To start solving for 𝐿 sub f,
let’s recall Newton’s second law of motion for rotation. Written this way, the second law
tells us that the torque on an object is equal to its change in angular momentum
over the time interval during which that torque was applied. And another way to write this is to
replace Δ𝐿 with the final angular momentum of the system minus the initial angular
momentum.

And now we see that this
relationship contains the variable we want to solve for, 𝐿 sub f. To rearrange so that 𝐿 sub f is
the subject of this equation, we can start by multiplying both sides by Δ𝑡,
canceling that factor out on the right. Next, what we do is add 𝐿 sub i,
the initial angular momentum of our system, to both sides, canceling that term on
the right. With the equation now in this form,
notice that in our particular example, we’re given values of 𝐿 sub i, 𝜏, and
Δ𝑡. And this means we can substitute
them in for their respective variables in this equation.

With these substitutions made,
before we add these terms together, let’s just check that the units of each one
agree with the units of the other. That is, we want to make sure that
a kilogram meter squared per second is equal to a newton times a meter times a
second. To check this, we can recall that a
newton is equal to a kilogram meter per second squared. When we substitute that in for our
unit of newton in the equation, we see that one factor of seconds cancels out
here. And if we gather all the units in
this term off to the right, we find that indeed they equal kilograms meter squared
per second.

All this tells us that we can add
this term to this term because they’re both in the same units. Now, if we multiply 1.8 by 15, the
result we get is 27. So our answer will be 12 kilograms
meter squared per second plus 27 kilograms meter squared per second. This is 39 kilograms meter squared
per second. And that’s the angular momentum of
our wheel after it accelerated.

As a side note, notice that this
amount of angular momentum was how much the angular momentum of the wheel changed
and that we needed to add this value to its initial value to find the true final
value of the wheel’s angular momentum.