### Video Transcript

In this video, we’re talking about
Newton’s second law of motion for rotation in terms of angular momentum and
time. Not only will we learn how to write
Newton’s second law of motion in these terms, but we’ll also see what we can learn
from graphs of the variables involved in this form of the second law.

As we get started, let’s recall
this familiar form of Newton’s second law of motion. The net force on an object is equal
to that object’s mass times its acceleration. If we recall further that an
object’s acceleration is defined as its change in velocity divided by a change in
time, then we can replace the acceleration 𝑎 in our form of Newton’s second law
with this fraction. And once we’ve done that, we can
further recall something about momentum. An object’s momentum, recall, is
equal to its mass multiplied by its velocity.

Now, if we take this equation for
momentum and look at the change in each side of this equation, Δ𝑝 and Δ𝑚 times 𝑣,
then in cases where the mass of our object is constant, we can write that the change
in an object’s momentum is equal to its mass times its change in velocity. We do all this because, looking
back at our form of Newton’s second law, we see here 𝑚 multiplied by Δ𝑣. By our equation for change in
momentum Δ𝑝, we can reexpress the second law this way. 𝐅 is equal to Δ𝑝 divided by
Δ𝑡. This is one way to write Newton’s
second law for motion that is linear.

It turns out though that it’s also
possible to write this law in what we can call a rotational version. To do this, we’ll take any
variables in this equation that apply to linear motion and we’ll replace them with
their rotational analogs. As a quick example of this, say
that we have an object moving in a circle. It’s moving with a linear speed 𝑣
a distance 𝑟 away from the circle’s center. In this case then, our object has
an angular speed — we call it 𝜔. For 𝜔 and 𝑣 then, both of these
variables refer to a speed, but one is linear and the other is rotational.

So then, looking back to this
equation, what we can call a linear expression of Newton’s second law of motion, the
variables here that are specifically linear are the force 𝐅 and a change in
momentum Δ𝑝. For us to write out a rotational
version of Newton’s second law of motion, we’ll need to replace these variables with
their rotational analogs. The rotational analog of force is
torque, symbolized using the Greek letter 𝜏. And then if we have some amount of
momentum 𝑝, the rotational analog of that is angular momentum 𝐿.

Going back to our object moving in
a circle, say that this object has a mass 𝑚. In that case, its linear momentum
would be 𝑚 times 𝑣, while its angular momentum would be 𝑚 times 𝑣 times 𝑟,
where 𝑟 is the perpendicular distance from the center of the circle to the tail of
the object’s velocity vector.

Knowing all this, let’s now make
these substitutions, 𝜏 in place of 𝐅 and 𝐿 in place of 𝑝, in our equation for
Newton’s second law. When we do this, we discover that
the torque on an object is equal to its change in angular momentum over the time
taken for that change to occur. And we see now that we’ve written
Newton’s second law of motion for rotation in terms of angular momentum and
time. Just as a side note, we didn’t have
to replace this variable Δ𝑡 because there’s no direction, either linear or
rotational, associated with time. That’s why we were able to carry it
over unchanged from our linear equation for Newton’s second law.

Now, we can understand this
equation a bit more clearly by looking at a graph of angular momentum 𝐿 versus time
𝑡. On this graph, we’ll want to
specify units for our variables. For the time 𝑡, we can use units
of seconds. And then to figure out what the
units of angular momentum are, let’s recall that 𝐿 is equal to 𝑚 times 𝑣 times
𝑟. Considering first mass, we know
that the SI base unit of mass is the kilogram. And then speeds are often recorded
in units of meters per second. And a radius may be reported in the
base unit of meters.

And so this tells us that, overall,
an angular momentum value has units of kilograms meter squared per second. So those are the units of 𝐿. And let’s say we have a graph of 𝐿
versus 𝑡 that looks like this. If we divide up this graph so that
in each of our two regions the slope or gradient of the curve is constant, then for
each one of the regions, we can calculate the torque that corresponds to these
changes in angular momentum over a change in time.

Let’s start out with our first
region, which corresponds to time values from zero up to six seconds. We can see that, over this time
span, there’s been a change in angular momentum. And that change is from zero up to
five kilograms meter squared per second. We can say then that for this
region, Δ𝐿 equals five kilograms meter squared per second and Δ𝑡, the change in
time, is six seconds. And our Newton’s second law for
rotation equation tells us that if we take the ratio Δ𝐿 to Δ𝑡, then that is equal
to the torque applied to some object to bring about this change in angular momentum
over this time. For this region of our graph, we
find that the torque is five-sixths kilograms meter squared per second squared. And since a kilogram meter per
second squared is equal to one newton, we see that this is equal to five-sixths of a
newton times a meter.

And then we can do something
similar on the second region of our graph. We once again look at the change in
time or the time interval. We see that’s from six up to 12
seconds. So Δ𝑡 is 12 seconds minus six
seconds, or simply six seconds. And then for Δ𝐿, we see we start
out at a value of five kilograms meters square per second but then end up at a value
of three kilograms meters squared per second. Subtracting the initial value from
the final value, this works out to negative two kilograms meter squared per
second. And so when we divide Δ𝐿 by Δ𝑡 to
solve for the torque applied to our object over this time interval, we find it’s
equal to negative one-third newton times a meter.

Now, looking at a graph of angular
momentum versus time is one way to better understand this relationship. But another way is to look at a
graph of torque versus time. What we’ve done here is kept our
horizontal axis the same, but we’ve changed our vertical axis so now it shows torque
rather than angular momentum. And let’s say that in this case,
for a different physical scenario than before, our graph looks like this.

Now, if we go back to this
equation, we can notice what will happen if we multiply both sides of the equation
by the time Δ𝑡. If we do that, we’ll get this
result that the torque 𝜏 times Δ𝑡 is equal to the change in angular momentum. This is a perfectly valid
restatement of Newton’s second law of motion for rotation. And this new form tells us that if
we take a torque value, shown on our graph on the vertical axis, and multiply it by
a time interval, shown on the horizontal axis, then we’ll wind up with a change in
angular momentum.

On our graph, this means that if we
take every value of the torque and multiply it by its corresponding time and add all
of those multiples together, then the sum of all that will be the total change in
angular momentum Δ𝐿. Graphically, we can calculate this
by solving for what’s called the area under the curve. This is the area between our graph
in the horizontal axis. If we find this area, then we’ll
have found Δ𝐿 because the two are equal.

So then, what is Δ𝐿 in the case of
this graph we have here? Well, our equation down here tells
us that we take torque and multiply it by a change in time. As we look at our graph, we see
that often the torque acting on our object is zero, but not all of the time. For this time interval here, from
two up to five seconds, the torque on our object is four newton meters. So in this equation for Δ𝐿, we’ll
use four newton meters for the torque 𝜏. And then for Δ𝑡, we’ll use our
time interval from two up to five seconds. That’s five seconds minus two
seconds, or simply three seconds. And so our overall change in
angular momentum is equal to 12 newton meter seconds. Or we could equivalently say 12
kilograms meter squared per second.

One important thing to keep in mind
here is that we’ve calculated a change in angular momentum. If our system already had some
nonzero amount, then its final angular momentum wouldn’t be this value we’ve
calculated but rather would be this plus its initial value. Knowing all this about Newton’s
second law of motion for rotation, let’s now get some practice with an example
exercise.

A spinning metal disk initially has
an angular momentum of 2.4 kilograms meter squared per second. A constant torque is applied to the
disk. Over a time of four seconds, its
angular momentum increases to 3.6 kilograms meter squared per second. What is the magnitude of the torque
that is applied to the disk?

So, in this example, we have a
spinning metal disk. And we’re told that, initially,
this disk has an angular momentum — we’ll call it 𝐿 sub i — of 2.4 kilograms meter
squared per second. Then we’re told that a constant
torque is applied to the disk and that, as a result, the disk’s angular momentum
increases to this new value. This means that the torque on the
disk was in the same direction as the disk was originally spinning.

Given that all this takes place
over a time interval of four seconds — we’ll label that value Δ𝑡 — we want to solve
for the magnitude of the torque that’s applied to the disk. So then we know the disk’s initial
angular momentum, its final angular momentum — we’ve called that 𝐿 sub f. We know that this change was due to
an applied torque and that this torque was applied for a time interval of four
seconds.

And now we can recall an equation
that relates these variables, angular momentum, torque, and time, sometimes called
Newton’s second law of motion for rotation. The torque on an object is equal to
its change in angular momentum divided by a change in time. In the case of our spinning metal
disk, the magnitude of the torque applied to it is equal to its final angular
momentum minus its initial angular momentum — that difference will equal Δ𝐿 — all
divided by the change in time Δ𝑡. Since we know the values of all
three of these variables, 𝐿 sub f, 𝐿 sub i, and Δ𝑡, we can substitute them in
now.

With these values plugged in, we
can see that in our numerator, we’re subtracting 2.4 kilograms meter squared per
second from 3.6 kilograms meters squared per second. That gives us 1.2 of those units of
angular momentum. And we divide this by four seconds,
giving us a value of 0.3 kilograms meter squared per second squared or equivalently
0.3 newtons times meters. This is the magnitude of the torque
applied to the disk over this time interval.

Let’s now look at a second example
exercise.

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.

Let’s now summarize what we’ve
learned about Newton’s second law of motion for rotation in terms of angular
momentum and time. In this lesson, we saw that we can
write Newton’s second law of motion in a rotational form using angular momentum and
time. We also saw that, given a graph of
angular momentum versus time, if we consider the gradient of that line, that is, the
change in angular momentum divided by the change in time, then we’re able to solve
for the torque applied using this form of Newton’s second law of motion. And lastly, we saw that when
instead we’re looking at a graph of torque versus time, we can solve for a system’s
change in angular momentum Δ𝐿 by multiplying the torque by the time interval, that
is, calculating the area under the curve.

This is a summary of Newton’s
second law of motion for rotation in terms of angular momentum and time.