- Good morning.
Today, we are going to continue our review of rotational
dynamics, as part of the AP Physics C Mechanics curriculum.
♫ Flipping Physics ♫
Bo, you wanted to talk about this last time,
so, here's your chance.
What is the equation for torque which includes
the cross product?
- Torque equals the cross product of F and r.
- Actually, Torque is the cross product of r and F.
- Does the order of r and F really matter?
- Well, yes, the cross product is not commutative.
- Of course it's not.
- And we should also point out that r is the position vector
from the axis of rotation to where on the object
the force is applied.
- And, remember, cross product is the area of
the parallelogram created by those two vectors.
For torque, it is the area with the sides r and F
that create the parallelogram.
And the magnitude of the torque is r times F
times the sine of the angle theta, which is the angle
between the direction of r and the direction of the force.
Now, let's do a basic example to review cross product
with unit vectors.
Vector A is negative I plus J minus 2K.
Vector B is 2i minus 3j plus 4K.
Billy, please work out A cross B.
- Well, the cross product is the determinant of a matrix.
This time, because we have three dimensions, X, Y and Z,
we have a three-by-three matrix.
So, the cross product is the determinant of the matrix
where the first row is I J K,
the second row is the components of A, or negative one,
one, negative two, and the third row is the components
of B, or two, negative three, four.
That is equal to the sum of the two-by-two determinants
for each of the unit vectors, although we subtract
the second determinant, the determinant that goes with J,
in this case.
So, cross out the row and column which has I in it,
and you get the determinant one, negative two,
negative three, four, times unit vector I.
Cross out the row and column which has J in and subtract
the determinants negative one, negative two, two, four,
times unit vector J.
Cross out the row and column which has K in it
and add the determinant negative one, one, two, negative three,
times unit vector K.
- [Bo] Mr. P, can we skip this step?
- Bo, I don't normally write out the step with the
two-by-two determinants in it.
However, I think it's a good idea to have done this
to see where the next step comes from.
Billy?
- Now we multiply one by four and subtract the product
of negative two and negative three, and multiply
that quantity by by unit vector I, then we subtract
the quantity negative one times four, minus the product
of negative two and two, all that times unit vector J,
and we add the product of negative one and negative three
and subtract one times two, all times unit vector K.
That works out to be the quantity four minus six times I
minus the quantity negative four plus four times J,
plus the quantity three minus two times K, and, actually,
all that we're left with is negative two I plus K.
- Thank you, Billy.
Bobby, please tell me about translational equilibrium.
- Well, we just talked about this during our
dynamics review.
An object is in translational equilibrium if the net force
acting on the object equals zero, and because net force
equals mass times acceleration, that means the object
is not accelerating.
- Very good.
And now, Bobby, please extend that
to rotational equilibrium.
- Okay, then, an object is in rotational equilibrium
when the net torque acting on the object equals zero.
This means, because net torque equals moment of inertia
times angular acceleration, that the object has no
angular acceleration, so the object is either rotating
with a constant angular velocity or is not rotating at all.
- Now, please remember, when you know an object is in
translational equilibrium and rotational equilibrium
about one axis of rotation, then you also know that object
is also in rotational equilibrium about any other
axis of rotation, which is why, for example, when we look at
a person standing on a ladder, we can sum the forces
in the X and Y directions and we can also sum the torques
about multiple axis of rotation.
Capital L stands for angular momentum,
and angular momentum is a vector.
In order to understand how angular momentum is related
to torque, we are going to start with Newton's Second Law.
Net force equals mass times acceleration is the linear form,
and net torque equals moment of inertia times
angular acceleration is the rotational form.
However, neither of those two use derivatives.
Net force equals the derivative of linear momentum
with respect for time is the linear form
of Newton's Second Law.
And net torque equals the derivative of angular momentum
with respect to time is the rotational form
of Newton's Second Law.
Where, of course, torque and angular momentum are vectors.
Bo, what is the equation for the angular momentum
of a particle?
- How can a particle have angular momentum?
- Any particle moving relative to an axis of rotation
can have angular momentum.
- Yeah, that's right.
It does not even have to be a particle, any object
which is moving relative to an axis of rotation
can have angular momentum.
- Bobby and Billy, you guys are correct.
And, Bo, we'll do an example at the end which shows
how a particle can have angular momentum.
Billy, why don't you take this one.
What is the equation for the angular momentum of a particle?
- The angular momentum for a particle or a rigid object
which is moving but not rotating equals the cross product
of r and momentum, and r is similar to what it is
in the torque equation.
r is the position vector from the location of
the center of mass of the moving object
to the axis of rotation.
- Then, the angular momentum equals r cross mass
times velocity, or r times mass times velocity
times the sine of the angle between the directions
of r and velocity?
- But, that is just the magnitude of the angular momentum.
The cross product equation is a vector
and includes direction.
If we use that last equation, we have to use
the right-hand rule to find the direction of torque,
just like we did in our last review.
Right, Mr. P?
- [Mr. P] That is correct, Bobby.
And, Bobby, what is the equation for the angular momentum
of a rigid object with shape which is rotating
around an axis of rotation?
- A rigid object with shape has an angular momentum
which equals the moment of inertia of that object
times that object's angular velocity.
- And the units for angular momentum are kilograms
times meters squared for moment of inertia,
times radians per second for angular velocity.
- But radians are dimensionless, and just a placeholder.
Therefore, the units for angular momentum
are kilogram meters squared per second.
- The derivation of the conservation of linear momentum
of a system is...
If the net external force acting on the system equals zero,
then the derivative of the momentum of the system
with respect to time is equal to zero.
That means the momentum of the system is not changing
as a function of time, and that is conservation
of linear momentum.
We can do the same thing to derive conservation
of angular momentum.
If the net external torque acting on the system is equal
to zero, then the derivative of the angular momentum
of the system with respect for time is equal to zero.
Therefore, the angular momentum of the system
is not changing as a function of time.
This is conservation of angular momentum.
One importance difference to remember here is,
that torque and angular momentum are relative
to an axis of rotation.
In other words, whenever you use conservation
of angular momentum, you have to identify
your axis of rotation.
Okay, let's now do an example problem which reviews
conservation of angular momentum, because...
We have to.
Bobby, could you please read the problem?
And Bo, could you please translate.
A piece of gum with mass, m, and velocity, v,
is spat at a solid cylinder of mass, M, radius, R,
and moment of inertia 1/2 MR squared.
The cylinder is on a horizontal axis through its center
of mass and is initially at rest.
- Stop.
Knowns for the gum are mass, little m,
and velocity initial, little v.
The knowns for the cylinder are mass, big M, radius, big R,
moment of inertia, which is 1/2, big mass times big radius
squared, and the cylinder has no initial angular velocity.
- Okay, so, I dislike using a lowercase m for the mass
of one object and an uppercase M for the mass
of a different object.
I know the AP writers like to do that sometimes.
However, to me, it just obfuscates what the masses
really represent.
So what I'm going to use is m sub g for the mass of the gum,
v sub g i for the velocity of the gum initial,
m sub c for the mass of the cylinder, and I'll just use
capital R for the radius of the cylinder
because it's the only object that has a radius.
And I do suggest that you do this, you clarify
your variables when you're solving an AP problem,
so that you can help the AP graders understand
what you're doing, which is always good.
Bobby?
- The line of action of the gum is located horizontally
at height y above the axis of the cylinder.
If the gum sticks to the cylinder, what is the final
angular velocity of the gum-cylinder system?
- Angular velocity final, which will be the same
for both the gum & the cylinder, equals question mark.
However, I don't really know what to do with y.
It's okay, Bo, it's actually a lot easier to describe
using a drawing.
Notice, because the gum sticks to the cylinder,
that we know y must be less than the radius, capital R,
of the cylinder.
Now, I know I said this was a conservation of angular
momentum problem.
However, can anyone explain to me how we know
angular momentum is conserved in this example?
Angular momentum is conserved in a system when
the net external torque acting on the system equals zero.
- When the gum hits the cylinder, there's a force on the gum
from the cylinder, which is directed radially outward.
- And because of Newton's Third Law, there is an equal
but opposite force acting on the cylinder from the gum,
which is directed radially inward.
- And if we sum the torques around the axis of rotation
of the cylinder, the torques from both of those forces
are zero, and therefore, the net external torque
acting on a system equals zero, and angular momentum
about the axis of rotation of the cylinder is conserved.
- Now, let's solve the problem.
Billy, could you please?
- Absolutely.
Conservation of angular momentum equals the initial
angular momentum of the gum, plus the initial
angular momentum of the cylinder, and that equals
the final angular momentum of the gum plus the final
angular momentum of the cylinder.
The gum is a particle, so we use the equation r
for the gum initial times the mass of the gum times
the velocity of the gum initial times the sine of theta
for the gum initial.
The cylinder is not rotating initially, so it has no
initial angular momentum.
The gum is still a particle after the collision,
so we use the same equation, only substitute in final
instead of initial.
The cylinder is a rotating object with shape after
the collision, so we use the moment of inertia of
the cylinder times the final angular velocity
of the cylinder.
But, I don't really know what to do from here.
This is where we need to find a relationship between r
for the gum initial, theta for the gum initial,
and this value y.
The issue is that both the r for the gum initial
and theta for the gum initial are changing as a function
of time as the gum approaches the cylinder.
The way we deal with this is we take the sine of theta
for the gum initial, which equals opposite over hypotenuse.
Opposite theta is this value y, and on the hypotenuse
is r for the gum initial, and we can solve for y,
it equals r for the gum initial times the sine of theta
for the gum initial.
So, while r for the gum initial and theta for the gum
initial both change as a function of position as that gum
approaches the cylinder, the value y, which equals
r for the gum initial times the sine of theta
for the gum initial is a constant.
y does not change as the gum approaches the cylinder.
So, we replace r for the gum initial times the sine
of theta for the gum initial with the value y.
Just so you know, this concept of replacing r times
sine theta with a constant value is pretty much guaranteed
to come up on the AP exam.
Bobby, what about the final values?
- The final equation for the gum is the same as the initial,
only replacing initial with final, so it should work out
to be the same thing, right?
- I don't think so.
Before the collision, the gum was moving linearly,
and after, it is moving in circular motion.
So, the equation is not going to work out the same way.
- Okay, so, because the gum is moving in a circle,
the r final for the gum is the radius of the cylinder,
and the velocity final for the gum is a tangential velocity,
which we know equals radius times angular velocity,
so we can substitute the radius of the cylinder
times the final angular velocity for the final velocity
of the gum.
- And theta for the gum final will be 90 degrees,
because tangential velocity is always tangent to the radius,
and the sine of 90 degrees is one.
- And we can substitute 1/2 times the mass of the cylinder
times the radius of the cylinder squared
for the moment of inertia of the cylinder.
- All we need to do now is factor out radius squared
times angular velocity final, and then divide by everything
on the right-hand side which is not angular velocity final,
and we have the angular velocity final equals y times
the mass of the gum times the velocity of the gum initial,
all divided by the radius of the cylinder squared
times the quantity, the mass of the gum plus the mass
of the cylinder divided by two.
- Now, when getting an answer like this which is all
variables, which is often going to happen on the AP Physics
C exam, it's good to just check your equation to make sure
it makes logical sense.
For example, Billy, does it make logical sense
that a larger velocity of the gum initial will result
in a larger final angular velocity of the system?
- Yeah, it makes sense to me that a larger initial velocity
of the gum would increase the final angular velocity
of the system.
How 'bout this?
According to our answer, a larger mass of the cylinder
will decrease the final angular velocity of the system.
That also makes sense to me.
- And a larger radius of the cylinder should decrease
the final angular velocity of the system.
That's logical, I think I get it.
- That is a good exercise to go through whenever you get
an answer that is all variables.
That brings us to the end of my review
of rotational dynamics.
Next, you can enjoy my linear versus rotational review,
or you could visit my AP Physics C review webpage.
Thank you very much for learning with me today,
I enjoyed learning with you.