- Good morning.
Today, we are going to review rotational dynamics
as a part of the AP Physics C Mechanics curriculum.
♫ Flipping Physics ♫
- Let's start by analyzing a rigid object with shape
which is rotating.
Every piece of a rotating object has kinetic energy,
and the total kinetic energy of the object
equals the sum of the kinetic energies
of each piece of the object.
In other words, we're treating this object
basically as a system of particles,
and the i here represents the total number of pieces
of the object, so i goes from one
to the total number of particles in the system of particles
or the total number of pieces of the object.
The equation for kinetic energy is one half
mass times velocity squared, and again,
we are adding the kinetic energies
for every particle in the system.
Because the object is rotating,
the velocity of every particle is a tangential velocity.
So we can substitute for the tangential velocity
the radius for each particle
times the angular velocity of each particle,
which are both squared because the tangential velocity is squared.
We can factor out the one half,
and we don't need an i on the angular velocity,
because every particle in this rigid object
has the same angular velocity.
We define this parenthetical quantity
as moment of inertia.
The moment of inertia of a system of particles
equals the sum of the quantity of the mass of each particle
times r squared, where r is the distance each particle is
from the axis of rotation of the system of particles.
Now that we have defined moment of inertia,
we know that the rotational kinetic energy
of a rigid object with shape, or a system of particles,
equals one half times the moment of inertia
times the angular velocity squared.
Bobby, what are the units for moment of inertia?
- Moment of inertia equals the sum of mass
times distance from axis rotation squared,
so kilograms times meters squared.
That's, that's really strange.
- I agree.
- If you compare with the equations
for linear kinetic energy to rotational kinetic energy,
you can see why moment of inertia
is also sometimes called rotational mass.
Right, Mr. P?
- Yes.
For some reason, kilograms times meters squared
seem strange to people for the units
for moment of inertia, and yes,
sometimes people call moment of inertia rotational mass.
Now let's take a look at the equation for
moment of inertia of a rigid object with shape
rather than a system of particles.
If we look at the limit of the equation
for the moment of inertia for a system of particles
and take the limit as the mass of each particle
becomes infinitesimally small,
we have the definition of an integral.
The moment of inertia of a rigid object with shape
equals the integral with respect to mass of r squared,
where r is defined as the distance
each infinitesimally small piece of the object is
from the object's axis of rotation.
Bo, which equation do you think
people often confuse this equation with?
- That looks a lot like the equation
for the position of the center of mass
of a rigid object with shape,
r center of mass equals the inverse of mass total
times the integral of r with respect to mass.
- Right, those two equations look very similar
and are often accidentally interchanged.
Please don't.
Let's do several moment of inertia derivations,
because we need to.
Let's start by deriving the moment of inertia
of a uniform, thin hoop about its cylindrical axis.
Billy, please describe what the words
uniform, thin, and cylindrical axis mean.
- Okay.
Uniform means the density of the hoop is constant.
Thin means the thickness of the hoop is small
compared to the radius of the hoop,
so we can consider the entire hoop
to be at distance capital R from its center,
and the cylindrical axis of the hoop
is the axis if you were to draw a line
for the axis normal to the whiteboard
right at the center of the hoop, which you have labeled z.
- Right, so we are determining the moment of inertia
of this hoop as it rotates around this point
in this plane like this.
It doesn't rotate this way, it rotates this way.
The moment of inertia about point z
is equal to the integral of r squared with respect to mass.
Notice that each piece dm
is located the same distance from point z.
It is the radius of the hoop, or capital R.
Because each piece dm is located the same distance capital R
from the axis of rotation,
we can replace the variable lowercase r
with a constant capital R
and take it out from underneath the integral.
The integral of just dm is mass.
Therefore, the moment of inertia of a uniform, thin hoop
about its cylindrical axis equals the mass of the hoop
times the radius of the hoop squared.
- That actually makes sense.
The mass of the entire hoop is located the same distance
capital R from the axis of rotation.
- Good.
Now, let's derive the moment of inertia
of a uniform, thin rod about its center of mass.
Bobby, could you please work on this one?
- Well, that is the moment of inertia about the y-axis,
which, again, equals the integral of r squared dm,
but I don't know what to do with that.
- I think you need to start with lambda,
the linear mass density which is constant
for the uniform rod.
- Yeah, it equals the total mass of the rod
divided by the total length of the rod,
but it also equals the mass of dm
divided by the length of dm in the x direction, which is dx.
- That, that doesn't help.
I don't know what to do.
- The goal here is to substitute in something for dm
which has dr in it, or actually dx, in this case,
because r is in the x direction.
If we multiply both sides by dx,
we have dm equals the linear mass density times dx,
and we can substitute in for the linear mass density
the total mass of the rod
divided by the total length of the rod.
Does that spark any memories?
- Sure.
Substitute mass over length times dx
in for dm in the moment of inertia equation.
The total mass of the rod and the length of the rod
are not dependent on position,
so mass over length can come out from under the integral,
and r is just a generic letter for direction,
so let's substitute in x instead,
which means that we can now put in
the limits for the integral.
X varies from negative one half of L
on the left side of the rod
to a positive one half L on the right side of the rod,
so those are our limits.
The integral of x squared with respect to x
is x cubed over three.
We can substitute in our limits
and cube both positive and negative L over two.
We now have L cubed over 24 plus L cubed over 24
in the brackets, which is 2L cubed over 24.
L cubed over L is just L squared.
So the moment of inertia of a uniform, rigid rod
about center of mass is 1/12 times the mass of the rod
times the length of the rod squared.
- Bobby, I'm glad you were able to remember that.
Billy, please determine the moment of inertia
of a uniform, rigid rod about one end.
So, it's the same object,
we simply moved the axis of rotation
from the center of mass to one end.
- Well, let's see.
It's the same object,
but we have moved the axis of rotation,
so it's actually the same integral
but with different limits.
So we can start from the middle of the previous example
with the moment of inertia about the y-axis
equals the total mass of the rod
divided by the length of the rod
times the integral of x squared with respect to x,
and the limits are now x initial is at the axis of rotation,
so x initial is zero.
X final is at the other end of the rod,
so x final is the total length of the rod, L.
Again, the integral of x squared with respect to x
is x cubed over three from zero to L.
Substituting in our limits gives us
mass over length times length cubed over three.
One of the Ls cancels out, and the moment of inertia
of a uniform, rigid rod about one end
equals one third times the mass of the rod
times the length of the rod squared.
- Now let's look at the parallel axis theorem.
If you know the moment of inertia of a rigid object
about its center of mass
and that rigid object has a uniform density,
you can figure out the moment of inertia
about some other axis of rotation
using the parallel axis theorem.
The moment of inertia about some other axis
is equal to the moment of inertia about the center of mass
plus the mass of the object times D squared,
where D is the distance between the new axis of rotation
and the center of mass of the object.
And be aware, this equation is not on the AP equation sheet.
For example, if we know the moment of inertia
of a uniform, rigid rod about its center of mass,
we can use that to determine the moment of inertia
about one of the ends of the rigid rod.
Using the parallel axis theorem,
we know the moment of inertia about one end of the rod
is equal to 1/12 times the mass of the rod
times the length of the rod squared
plus the mass of the rod
times the distance between the new axis of rotation
and the center of the mass of the rod squared,
which happens to be half the length of the rod.
Squaring L over two gives us one fourth mL squared,
and factoring out mass times length squared gives us
4/12, or one third, times a total mass of the rod
times the total length of the rod squared.
- [Bo] Didn't we already derive that?
- True, but it gave us a chance to confirm our answer
and review the parallel axis theorem.
How about you do an example where we have to use
the parallel axis theorem?
Bo, please determine the moment of inertia
of a uniform, thin hoop about its rim.
- Sure.
Using the parallel axis theorem,
we already know the moment of inertia
of a uniform, thin hoop about its cylindrical axis
is mass times the radius squared,
so we add the mass of the hoop
times the square of the distance from the center of the hoop
to the rim of the hoop,
which is just the radius capital R of the hoop.
So the moment of inertia of a uniform, thin hoop
about its rim is two times its mass
times the radius of the hoop squared.
- Great.
Next up is the concept of torque.
The symbol for torque is a lowercase Greek letter tau,
and it equals r F sine theta,
but what are r, F, and theta?
Bobby?
- Okay, let's see.
r is the distance from the axis of rotation
to where on the object the force is applied,
F is the magnitude of the force causing the torque,
and theta is the angel between r and F.
- We should point out torque is a vector,
and that equation is only the magnitude of the torque.
- I thought torque was the cross-product of r and F.
- We will get to the cross-product equation for torque
in our next review session, but not yet, Bo.
If we look at this top view
of a force being applied to a door at a distance r
from the axis of rotation of the door,
you can see we have this distance d,
which we can solve for using the equation for sine.
Sine of theta equals opposite over hypotenuse,
or d over r, and therefore, d is equal to
r times the sine of theta.
d is defined as the moment arm or the lever arm,
or I've even seen it called the effective distance.
Notice that if d or r sine theta is equal to zero,
then no torque is being applied to the door.
Or if the moment arm or lever arm is maximized,
using an angle of 90 degrees,
then the maximum torque possible from a force
is being applied to the door.
Billy, what are the units for torque?
- Looking at the equation for torque,
torque must be in newton meters, which is joules.
- Joules are energy.
Torque is in newton meters.
- Aren't they the same thing?
- Yes, torque is in newton meters,
which technically are the same thing as joules.
However, to help differentiate between torque and energy,
we use newton meters for torque and joules for energy.
- [Billy] But what is torque?
- That's a good question, Billy.
Does anybody have an answer?
- Isn't torque like a rotational force?
- Force is the ability to cause an acceleration
of an object, so-
- So torque is the ability of a force
to cause an angular acceleration of an object.
- Okay, that makes sense.
A net torque causes an angular acceleration of an object.
It's like a rotational force, yeah.
- That is a very nice segue, thank you.
- What does Dean Kamen have to do with this?
- No, segue, not Segway.
- Yeah. - Oh.
- Let's talk about the rotational form
of Newton's second law.
Newton's second law is the net force
equals mass times acceleration
where both force and acceleration are vectors,
and the rotational form of Newton's second law
simply replaces each of those variables
with its rotational equivalent.
Net torque equals moment of inertia
times angular acceleration,
where torque and angular acceleration are vectors.
Bo, what must you identify every time you sum the torques?
- When summing the torques, you always have to identify
the objects, the axis of rotation,
and the positive direction.
- And in order to sum the torques on multiple objects,
the angular acceleration of each object
around the same axis of rotation must be the same.
- Speaking of direction, actually, no, wait.
Before we talk about the direction of torque,
now that we've defined moment of inertia,
pulleys can have mass, and when pulleys have mass,
the force of tension on either side of a pulley
is no longer the same.
Let me repeat that.
When a pulley is massless and frictionless,
the forces of tension on either side of the pulley
are the same.
However, when a pulley does have mass,
those forces of tension on either side of a pulley
are no longer the same.
Okay, back to direction.
Class, what handy tool do we use
in order to find the direction of torque?
- The right-hand rule.
- That's right, the right-hand rule.
- Don't be too cool for the right-hand rule.
- That's right, we need to limber up.
- We are so not cool.
- Yup, please do not be too cool for the right-hand rule.
Limber up and be willing to rotate your shoulders
and your hips and find your right hand.
- Right hand.
- Start with the fingers at the axis of rotation.
Point them along the direction of r.
Curl your fingers in the direction of the force,
and your thumb points in the direction of the torque,
which is, in this case, into the board.
Works from the other side, as well.
Start with your fingers at the axis of rotation.
Point them along the distance of r.
Curl your fingers in the direction of the force.
Stick out your thumb, and your thumb points
in the direction of the torque.
- Fingers of your right hand start at the axis of rotation.
Point your fingers along r.
Curl your fingers in the direction of the force,
and your thumb points in the direction of the torque.
- In this example, the torque is into the board,
which is, by definition, negative.
Out of the board is positive,
and into the board is negative by definition.
However, it is always a good idea
to identify the positive direction of torque.
Now let's talk about rolling without slipping.
This is when an object is rolling along a surface
and not slipping.
The equations which indicate
when an object is rolling without slipping are
the velocity of the center of mass of the object
is equal to the radius of the object
times the angular velocity of the object,
and the acceleration of the center mass of an object
is equal to the radius of the object
times the angular acceleration of the object.
Notice how similar these two equations are
to two tangential equations.
I bring this up
because the only one of these four equations
which is on the AP equation sheet
is the equation tangential velocity
equals the radius times the angular velocity.
You have to memorize the other three.
Now, just to be clear, the equations which describe
rolling with slipping are
that the two equations for rolling without slipping
are not true.
Remember, when an object is rolling,
it has both translational and rotational kinetic energy.
That completes the first part of my review
of rotational dynamics.
Next, feel free to enjoy the second part
of my review of rotational dynamics
or visit my AP Physics C review web page.
Thank you very much for learning with me today.
I enjoyed learning with you.