Video Transcript
In this video, we’re going to learn
about angular momentum. We’ll learn what this term means,
how it relates to linear momentum, and how to work with it practically.
To get started, imagine that you’re
the owner of a wind farm which uses wind power generators to make electricity. Today is the day of the company’s
picnic. And since all the employees will be
at the picnic with no one able to monitor and attend to the wind farm, per safety
regulations you’ll need to shut the farm down while it’s unattended. With the wind turbines currently
rotating as they usually do in the wind, you’d like to know just how hard it will be
to bring them to rest. To figure this out, it will be
helpful to know something about angular momentum.
To get started talking about
angular momentum, let’s imagine a planetary orbit scenario where a smaller planet
orbits a larger one in a circular path. If we were to write down some of
what we know about this scenario, we know that this smaller planet has some mass we
can call 𝑚. And we also know that at the
instant we’ve shown it, it has a tangential velocity we can call 𝑣. Considering just these two terms,
we see that if we combine them, we can form the linear momentum of this rotating
planet.
We’ve seen before that linear
momentum 𝑝 is equal to an object’s mass times its velocity. Since the orbiting planet is also
moving in a circular path, we can draw a vector that goes from the point around
which it moves to the planet itself. And we’ll call that vector the
distance vector 𝑟.
By adding in this distance vector,
which, we can see, will change as the position of the planet in its circular orbit
changes, we’ve introduced an angular, or rotational, aspect to the motion of this
orbiting planet. After all, if we draw in the
position of the planet at some later time, we can see that’s it’s moved through an
angular distance we can call 𝜃 during that time.
When we combine these terms we’ve
used to get linear momentum with the distance of vector 𝑟, that’s when we arrive at
an angular momentum. Angular momentum is symbolized by a
capital letter 𝐿, and it’s a vector quantity with both magnitude and direction. Angular momentum is equal to a
cross product of our distance vector 𝑟 — remember, this vector points from the
point of rotation to the object that is doing the rotating — and the linear momentum
of that rotating object 𝑝.
Since linear momentum 𝑝 is equal
to 𝑚 times 𝑣, we can also write our angular momentum as 𝑟 cross 𝑚 times 𝑣, or
since mass is a scalar quantity 𝑚 times the quality 𝑟 cross linear velocity
𝑣. With this new quantity we’re
calling angular momentum, we’d like to know what’s useful about it or why we’d be
interested in an object’s angular momentum.
The great thing about angular
momentum is, as we’ll see in a later segment, it’s a conserved quantity just like
linear momentum 𝑝. That means that over any physical
interaction or process, we know that the angular momentum at the beginning is the
same as the angular momentum at the end. We’ve said that angular momentum
has to do with rotation. And in fact, we can write the
expression for angular momentum in terms of rotational variables.
Moving from speaking of vectors to
speaking of scalars, we know that linear speed 𝑣 is equal to the radial distance an
object moves from its axis of rotation multiplied by its angular speed 𝜔. If we assume that the vectors 𝑟
and 𝑣 are perpendicular to one another, then we can write that the magnitude of 𝐿
is equal to 𝑚 times 𝑟 times 𝑣 using simple multiplication. Then, we can substitute 𝑟𝜔 in for
𝑣 so that now 𝐿 is equal to 𝑚𝑟 squared 𝜔.
And if we look at 𝑚 times 𝑟
squared, we may recall that this is the expression for the moment of inertia of a
point mass. Under the condition of working with
a point mass then, where its velocity and radial distance are perpendicular to one
another, we can say that the magnitude of angular momentum is equal to moment of
inertia times 𝜔.
Let’s take a moment at this point
and consider 𝜔 as a vector quantity, that is as angular velocity. Over here on our diagram of the
orbiting planet, if we were to draw in 𝜔 the angular speed of the planet, we might
do it with a curving arrow like this to show that the planet is moving in a
counterclockwise direction in a circle. Here is something that’s
interesting though. If we consider this 𝜔 as a vector,
that is as an angular velocity, then the actual direction of 𝜔 under these
circumstances is out of the page at us.
That might seem strange like it
contradicts the direction of the arrow that we drew showing the planet’s motion
direction. But going back to our convention
we’ve sometimes used before of a right-hand rule, if we rotate our fingers in the
direction of the planet’s motion, that is, in some sense, the direction of its
angular position, then our thumb points in the direction of 𝜔, the angular
velocity. That’s why this counterclockwise
rotating planet has an angular velocity that points out of the page at us.
The reason we say all this is
because we can actually go back to our equation for angular momentum in terms of
rotational variables, and we can write in 𝜔 as a vector and 𝐿 as a vector, and
this equation is correct as written. That’s to say, the direction of an
object’s angular momentum is the same as the direction of its angular velocity, just
like we saw for linear momentum 𝑝 matching up with the direction of linear velocity
𝑣.
Let’s consider one more thing about
angular momentum. What if we were interested not in
the angular momentum at a point in time but in a change in angular momentum Δ𝐿. To figure out what this quantity
might be, let’s go back to the linear analog of angular momentum, that is linear
momentum 𝑝.
We can recall that there’s a
theorem with the long name, the impulse-momentum theorem. This theorem says that the change
in an object’s momentum Δ𝑝 is equal to the net force acting on that object times
the time over which this average force acts. Now knowing that linear variables
such as these typically have corresponding angular or rotational variables, how
would we rewrite the impulse-momentum theorem for angular momentum rather than
linear momentum?
We would replace Δ𝑝 with Δ𝐿, and
that would equal not force but angular force, which is torque, multiplied by
Δ𝑡. Time is the same regardless of
whether we’re talking about rotational or linear motion. This relationship, which we could
call the rotational analogue of the impulse-momentum theorem, is an accurate
statement of changes in angular momentum Δ𝐿. Realize that the torque in this
expression 𝜏 is a net average torque. We assume that it’s constant over
the time Δ𝑡 that it acts.
Now that we have a sense for what
angular momentum is in terms of linear and rotational variables and we also have a
feel for how it changes and what it depends on, let’s get some practice with these
ideas through an example.
Three particles move independently
of each other, and each particle is subject to a force perpendicular to the
direction of its instantaneous velocity. What is the total angular momentum
about the origin of the particles? What is the rate of change of the
total angular momentum about the origin of the particles?
We can call the total angular
momentum of the particles capital 𝐿 and the rate of change of that total Δ𝐿 over
Δ𝑡. To start off solving for 𝐿, the
total angular momentum of our three particles 𝑚 one, 𝑚 two, and 𝑚 three about the
origin, we can recall the mathematical definition for angular momentum. 𝐿 as a vector is equal to its
position vector crossed with its momentum. And knowing that 𝑝 momentum is
equal to mass times velocity, we can also write this as 𝑚 times the quantity 𝑟
cross 𝑣.
In our situation, we have three
masses at different locations, and, therefore, each one has its own different
position vector. We can write out what those
position vectors are by observing our mass positions relative to the origin. We see that 𝑚 one is at a position
of negative two meters on the 𝑥-axis and positive one meter on the 𝑦. So, we write its position vector as
negative two 𝑖 plus 𝑗 meters.
The position of mass 𝑚 two
relative to the origin is plus four meters in the 𝑥-direction and plus one meter in
the 𝑦. So, it’s position vector is four 𝑖
plus 𝑗 meters. And 𝑚 three is at a position of
positive two meters in the 𝑥-direction and negative two meters in the 𝑦-direction
relative to the origin. So, we can write its position
vector, 𝑟 three, as two 𝑖 minus two 𝑗 meters.
Since we’re given the velocities
and masses of each one of our three masses, we now have all the information we need
to calculate the angular momentum of each one. Since angular momentum 𝐿 is a
vector and we’ll take a cross product to compute that vector, let’s remind ourselves
of the rule for crossing two different vectors, we’ll call them generically 𝐴 and
𝐵, and assume they’re three-dimensional.
Given two three-dimensional vectors
𝐴 and 𝐵, their cross product is equal to the determinant of the three-by-three
matrix where the first row of the matrix is the unit vectors 𝑖, 𝑗, and 𝑘 of our
three-dimensional space. The last two rows are the 𝑖, 𝑗
and 𝑘 components of these two vectors 𝐴 and 𝐵 respectively. We’ll use this rule to calculate
the angular momentum of mass one, we’ll call it 𝐿 one, and also the same for mass
two, we’ll call it 𝐿 two, and the same thing for mass three. Then, we’ll add together those
three angular momenta to solve for 𝐿, the total.
Starting with 𝐿 one, that’s equal
to 𝑚 one times 𝑟 one cross 𝑣 one. We know 𝑚 one, the mass, and 𝑣
one, the velocity, from our diagram. And we’ve solved for 𝑟 one, the
position vector of that mass relative to the origin. Plugging in for our value 𝑚 one
and following our cross-product rule for setting up the relationship between 𝑟 one
and 𝑣 one, we know that because 𝑟 one and 𝑣 one only have 𝑥- and 𝑦-components,
that means the direction of their cross product will only have a 𝑧-component. Therefore, 𝐿 one will only have a
𝑘-hat direction.
This means that when we evaluate
this matrix, we only need to evaluate the 𝑘th component because the 𝑖- and
𝑗-components will be zero. When we make that evaluation, we
take negative two meters and multiply it by 4.0 meters per second and subtract from
that one meters times zero, which is zero. Overall then, 𝐿 one is 2.0
kilograms times negative 8.0 meters squared per second in the 𝑘-direction, or
negative 16 kilograms meter squared per second in the 𝑘-direction.
Next, we move to calculating 𝐿
two, the same value but for 𝑚 two, 𝑟 two, and 𝑣 two. Again calculating only the
𝑘-component because that’s the only nonzero component to 𝐿 two, we find it’s equal
to 4.0 kilograms times negative 5.0 meters squared per second in the 𝑘-direction,
or negative 20 kilograms meter squared per second 𝑘. Then, we do the same thing for 𝐿
three with 𝑚 three, 𝑟 three, and 𝑣 three. This vector comes out to 6.0
kilograms meter squared per second again in the 𝑘-direction.
Now that we’ve solved for 𝐿 three,
𝐿 two, and 𝐿 one, we’re able to add all these three together and solve for the
overall angular momentum 𝐿. This is equal to negative 16 minus
20 plus 6.0 kilograms meter squared per second in the 𝑘-direction, or negative 30
kilograms meter squared per second in the 𝑘-direction. That’s the total angular momentum
of this system of masses about the origin.
Next, we wanna solve for the time
rate of change of that total angular momentum. We recall from the rotational
version of the impulse-momentum theorem that the change in angular momentum 𝐿 is
equal to the total average torque 𝜏 multiplied by Δ𝑡. This tells us that Δ𝐿 over Δ𝑡 is
equal to the total torque 𝜏. Each one of the three masses
experiences a torque. So, to solve for the overall torque
𝜏, we’ll need to solve for the individual torques and sum them.
Since torque is equal to 𝑟 cross
𝐹, and we’ve already solved for the position vectors of our three masses and are
given their forces, we can use the same cross product rule to solve for the overall
torque. Just like with our angular
momentum, since all of our position vectors and all of the forces acting on our
particles have only 𝑥- and 𝑦-components, that means all of our torques 𝜏 one, 𝜏
two, and 𝜏 three will only have 𝑘-components. And that means we only need to
evaluate the 𝑘-component of our three-by-three matrix.
In the case of 𝜏 one, plugging in
for 𝑟 one and 𝐹 one by component, this is equal to 6.0 newton meters in the
𝑘-direction. Moving onto 𝜏 two, the torque on
particle two, this is equal to the cross product of 𝑟 two and 𝐹 two, or 40 newton
meters in the 𝑘-direction. And lastly, we calculate 𝜏 three,
plugging in for 𝑟 three and 𝐹 three in our matrix. This gives us a result of negative
16 newton meters in the 𝑘-direction.
With values now for 𝜏 three, 𝜏
two, and 𝜏 one, we can add these together to solve for the overall torque 𝜏. This is equal to 6.0 plus 40 minus
16 newton meters in the 𝑘-hat direction, or 30 newton meters in the 𝑘-direction. And since this is the overall
torque acting on our system of masses, by the impulse-momentum theorem, it’s also
equal to the time rate of change of angular momentum.
Let’s summarize what we’ve learned
about angular momentum. We’ve seen that angular momentum is
the vector product of an object’s position vector with its linear momentum. As an equation, 𝐿 is equal to 𝑟
cross 𝑝. We’ve also seen that angular
momentum can be written in terms of rotational variables, that it’s equal to an
object’s moment of inertia times its angular velocity. And finally, we’ve seen there is a
rotational version of the impulse-momentum theorem. It says the change in an object
angular momentum equals the torque acting on that object times the time over which
it acts.
