- [Instructor] The rotational kinematic formulas
allow us to relate the five different
rotational motion variables
and they look just like the regular kinematic formulas
except instead of displacement,
there's angular displacement.
Instead of initial velocity
there's initial angular velocity.
Instead of final velocity
there's final angular velocity.
Instead of acceleration there's angular acceleration
and the time is still just the time.
You only get the first two of these
on the AP exam formula sheet.
You do not get three and four.
And just like the regular kinematic formulas,
these rotational kinematic formulas
are only true if the angular acceleration is constant.
What do each of these rotational variables mean?
Well, the angular displacement is the amount of
angle the object has rotated through
in the certain amount of time t.
Angular velocity is defined to be the amount of
angle you rotated through per time
just like regular velocity is the displacement per time,
and the angular acceleration is defined
to be the amount of change in angular velocity per time.
Just like regular acceleration is the change
in regular velocity per time.
For something rotating in a circle,
technically the angular velocity points
perpendicular to that plane of rotation
but it's easiest to just think of omega
as being counterclockwise or clockwise.
And there's relationships between these
angular variables and their linear counterparts.
To get the arc length s the object has traveled through
you just multiply the radius of the path
by the amount of angular displacement.
To get the speed of the object
just multiply the radius of the path
by the angular speed of the object.
To get the tangential acceleration
multiply the radius of the path
by the angular acceleration.
Note that this is the tangential acceleration.
This acceleration causes the object
to speed up or slow down.
It's the centripetal component of the acceleration
that causes the object to change directions
and the formula for that is still
just v squared over r.
If an object's moving in a circle
it must have centripetal acceleration
because it's changing directions
but only if it's speeding up or slowing down
will it have tangential acceleration
and angular acceleration.
What's an example problem involving
the angular motion variables look like?
Let's say an object is rotating in a circle
at a constant rate.
Which would best describe
the three different types of accelerations of the object?
Well, if an object's moving in a circle at all
there has to be centripetal acceleration
so that's got to be non-zero.
And if it's rotating at a constant rate
there's no change in omega
and that's means the angular acceleration is gonna be zero.
If the angular acceleration is zero
the tangential acceleration would also be zero.
Only when the object is speeding up or slowing down
do you have angular acceleration
and tangential acceleration.
This change the speed
and centripetal acceleration changes the direction.
What does torque mean?
Just like force is what causes acceleration,
torque is what causes angular acceleration.
In order for an object to speed up or slow down
in its angular motion,
there's got to be a net torque on the object.
What causes a torque?
Forces cause torque.
In order to have a torque you have to have a force
but the same force could exert a different torque
depending on where that force is exerted.
If the force is exerted far from the axis of rotation,
you'll get more torque for that given amount of force
compared to forces that are exerted
near the axis of rotation.
This r represents how far that force is applied
from the axis.
And to maximize that force
you would actually wanna point it perpendicular to this r
since sine of 90 degrees is equal to one.
In other words, to maximize the amount of torque you get
exert the force as far away as possible from the axis
and exert that force perpendicular
to the line from the axis to that force.
You might get many angles in a problem
but this angle here is always the angle
between the r and the F.
And just like an object is in translational equilibrium
if the net force is zero,
we say that an object is in rotational equilibrium
if the net torque is zero.
This would cause the angular acceleration to be zero
just like translational equilibrium
causes the acceleration to be zero.
Torque is a vector so it has a direction.
Typically it's easiest to think of the direction
as just being counterclockwise or clockwise
based on which way that force
would cause the object to rotate.
And since torque is r times F,
the units are gonna be meters times newton or newton meters.
What's an example problem involving torque look like?
Let's say you had this rod with this axis here
and there were forces applied as shown.
We wanna know how large would the force F have to be
in order for this rod to be in rotational equilibrium.
Remember, rotational equilibrium means
that the net torque is equal to zero.
In other words, all the torque that's pointing clockwise
would have to equal all the torque
that points counterclockwise
in order for the system to be balanced.
The three newton force and the one newton force,
you're trying to rotate this system clockwise
and the unknown force F is trying to rotate
the system counterclockwise.
This green one newton force
isn't actually exerting any torque
even though the r value is not zero.
The angle between the force and the r value
is gonna be 180 degrees
and the sine of 180 degrees is zero.
Which makes sense since this force
isn't actually causing this rod to rotate
clockwise or counterclockwise.
The torque and the clockwise direction would be
one meter times three newtons
to find the torque from the three newton force.
Plus you wouldn't use two meters
for the r of the one newton force.
You have to find the r from the axis
which is gonna be three meters times one newton force
which gives a total clockwise torque of six newton meters,
and we can write the torque
applied by the unknown force F as one meter times F.
In order for six newton meters to equal one times F,
the force F just has to equal six newtons.
What's rotational inertia mean?
Well, an object with a large rotational inertia
will be hard to get rotating and harder to stop rotating.
Basically the rotational inertia tells you
how much an object will resist angular acceleration.
Just like regular inertia tells you
how much an object will resist regular acceleration.
And this rotational inertia is often referred to
as the Moment of inertia.
How do you make the rotational inertia large?
Well you can increase the rotational inertia
if you place the mass far from the axis of rotation
and you can make the rotational inertia smaller
if you place the mass close to the axis of rotation.
In other words if you could push the mass
closer to the axis of rotation
which is the point about which the object rotates,
you can make the moment of inertia smaller and smaller.
To find the moment of inertia or rotational inertia
of an object whose entire mass rotates
at the same radius r,
you can just use the formula I
equals the mass that's rotating,
times how far it's rotating from the axis squared.
This formula's not given, you have to memorize it.
I equals mr squared.
And if you had many masses rotating at different r's
you could just add up all the contributions
from each single mass.
Now if you have a continuous object
whose mass is not all at the same radius from the axis,
the formulas are a little more complicated.
For a rod rotating about its center,
the moment of inertia would be 1/12
the mass of the rod times
the entire length of the rod squared.
For rod rotating about one end,
the moment of inertia is gonna be larger
since more mass is distributed farther from the axis
and this formula is 1/3 the mass of the rod
times the entire length of the rod squared.
The rotational inertia of a sphere
rotating about an axis through its center
would be 2/5 the mass of the sphere
times the radius of the sphere squared.
And the rotational inertia of a cylinder or a disk
rotating about an axis through its center
would be 1/2 the mass of the disk
times the radius of that disk squared.
Another example that comes up a lot
that you wouldn't be given the formula for is a hoop.
That is to say where all the mass is distributed
around the center point with a hollow center.
Since all the mass is at the same radius r,
the formula for the rotational inertia of a hoop
is just the same as the formula for the rotational inertia
of a single mass rotating at radius r.
The fact that the mass is distributed in a circle
doesn't actually matter since the mass
still stayed at the same radius r away.
Rotational inertia is not a vector
so it's always positive or zero
and the units since it's mr squared would be
kilograms times meter squared.
What's an example problem involving
rotational inertia look like?
Let's say two cylinders are allowed to roll
without slipping from rest down a hill.
The mass of cylinder A is distributed
evenly throughout the entire cylinder.
Cylinder B is made from a more dense material
and it has a hollow center
with the mass distributed around that hollow center.
If the masses and radii of the cylinders are the same,
which cylinder would reach the bottom of the hill first?
To figure out which cylinder gets
to the bottom of the hill first
we have to ask which one would roll more readily.
The cylinder with the least moment of inertia
is gonna be easier to rotate.
That means it would roll more readily
and get to the bottom of the hill faster.
Whenever the mass is distributed farther away from the axis,
the object's gonna have a larger moment of inertia
so since the mass of cylinder B overall
is farther away from the axis compared to cylinder A,
cylinder B has a larger moment of inertia,
that means it's harder to rotate.
It will take longer to get down the hill
and cylinder A is gonna win.
What's the angular version of the Newton's Second Law?
Well, Newton's Second Law says
that the acceleration is equal to the net force
divided by the mass
and the angular version of Newton's Second Law says
that the angular acceleration is equal to the net torque
divided by the rotational inertia.
M tells you how much an object resist acceleration
and the moment of inertia or rotational inertia
tells you how much an object resists angular acceleration.
Just like when you add up force vectors
you have to be careful with positive and negative signs.
The same holds true with the torque vectors.
You've got to treat either counterclockwise
or clockwise as positive
and then be consistent with it.
What's an example problem involving
the angular version of Newton's Second Law look like?
Let's say the rod shown below
has a rotational inertia of two kilogram meter squared
and has the forces acting on it as shown.
We wanna know what the magnitude
of the angular acceleration is of the rod.
We use Newton's Second Law in angular form
which says that the angular acceleration is the net torque
divided by the rotational inertia.
We got the rotational inertia,
we just need the net torque.
We have to figure out the total torque
from all these forces.
The torque from the one newton force
would be r which is gonna be three meters
from the axis to that force one newton.
And since it's applied perpendicular,
the sine of 90 is gonna be one.
The torque from the one newton force
would be three newton meters
in the counterclockwise direction.
And the torque from the four newton force
would be one meter
since it's applied one meter from the axis
times four newtons and we get four newton meters
in the clockwise direction.
That means the total net torque
when you have four newton meters in the clockwise direction
and three newton meters in the counterclockwise direction
would just be one newton meter
in the clockwise direction
since four is one unit bigger than three.
And now we divide by the rotational inertia
which was two
which gives us an angular acceleration of 1/2 or 0.5.
What's rotational kinetic energy mean?
Well if an object is rotating or spinning,
we say it has rotational kinetic energy.
If the center of mass of an object is moving
and the object is rotating,
we typically say that object has
translational kinetic energy and rotational kinetic energy.
They're both kinetic energies,
this is just a convenient way to delineate
between two types of kinetic energy
and a particularly convenient way
to find the total kinetic energy
for something that's moving and rotating.
The formula for rotational kinetic energy
is 1/2 times the moment of inertia
or the rotational inertia times the angular speed squared.
Which makes sense because the formula
for regular kinetic energy is 1/2 times the regular inertia,
the mass times the regular speed squared.
Again, if an object is rotating,
it's got rotational kinetic energy.
If the center of mass of an object is moving
it's got regular translational kinetic energy.
And if the center of mass is moving
and the object is rotating,
then we say that object has both rotational energy
and translational energy.
Rotational kinetic energy is not a vector.
It is always positive or zero
and the units can be written as
kilogram meter squared per second squared
but it's an energy so we know that just has to equal joules.
What's an example problem
involving rotational kinetic energy look like?
Let's say a constant torque is exerted on a cylinder
that's initially at rest
and can rotate about an axis through its center.
Which of these curves would best give
the rotational kinetic energy of the cylinder
as a function of time?
Well if there's a constant amount
of torque on an object,
that will cause a constant angular acceleration.
And if the angular acceleration is constant,
we can use the kinematic formulas
to figure out the final velocity of this object.
The final angular velocity if it's started at rest
would just be alpha times t.
That means the rotational kinetic energy of this object
could be written as 1/2 the moment of inertia
which is a constant times omega squared.
Which in this case would be 1/2 I
times alpha t squared.
Since the function for kinetic energy
is proportional to the time squared,
if you graph kinetic energy as a function of time,
it would look like a parabola
so the correct answer would be B.
What's angular momentum?
Well, the reason we care about angular momentum
is that it will be conserved for a system
if there's no external torque on that system.
And just like regular momentum is mass times velocity,
angular momentum will be the rotational inertia
times the angular velocity.
And this is a convenient formula
to find the angular momentum of an extended object
whose mass is distributed at different points
away from the axis of rotation.
The strange thing about angular momentum
is that even a point mass moving in a straight line
can have angular momentum.
To find the angular momentum of a point mass
moving in a straight line,
take the mass of the object
times the velocity of that object,
and either multiply by how far that object is from the axis
times sine of the angle between
the velocity vector and that R.
Or the easier way to do it is to just multiply by
the distance of closest approach
which is how close that mass will ever get to
or ever has been from the axis.
In other words to determine the angular momentum
of this mass moving in a straight line,
draw a straight line along its trajectory
and ask how close has it gotten
or ever will get to the axis?
That's the capital R I'm talking about.
And if you take that times the mass times velocity,
you'll get the angular momentum of that point mass.
Angular momentum is a vector
and it's easiest to just think about the direction
of angular momentum as being
either counterclockwise or clockwise
depending on which way the object is rotating.
And as for the units if you multiply mass
of kilograms times meters per second times meters,
you'd get kilogram meter squared per seconds
as the units of angular momentum.
What's an example problem
involving angular momentum look like?
Let's say a clay sphere of mass M
was heading toward a rod of mass three M and length L
with a speed v.
The rod is free to rotate about an axis around its end.
If the clay sticks to the end of the rod,
what would be the angular velocity of the rod
after the clay sticks to the rod?
And we're given that the moment of inertia
of a rod about its end is 1/3 ML squared.
Since there's gonna be no
net external torque on this system,
the angular momentum of this system is gonna be conserved.
The only object in this system
that has angular momentum initially is this clay sphere.
Since this is a point mass moving in a straight line
we'll use the formula M times the velocity,
times the closest it will ever get to the axis which is L,
the length of the rod.
That's gonna have to equal the final angular momentum
which we could write as I times omega.
And this I would be the moment of inertia
of both the rod and the clay
that is now stuck to the end of the rod.
We'd have MvL equals the total moment of inertia,
moment of inertia of the rod is 1/3 mass of the rod
which is three M times the length of the rod squared.
Plus the moment of inertia of this piece of clay
stuck to the end of the rod
rotating in a circle is gonna be
the mass of the clay times the radius of the circle
that clay traces out which is the length of the rod.
In other words, we're using the formula
for the moment of inertia of a point mass
whose entire mass is rotating
at the same radius from the center.
And we add that to the moment of inertia
of the rod itself.
We multiply by omega.
The term in brackets comes out to be two ML squared.
We can cancel the M's,
we can cancel one of the L's
and we get that omega is gonna equal v over two L.
The last topic I wanna talk about
is the more general formula
for the gravitational potential energy.
Why do we need a more general formula?
Well, if you're in a region
where the gravitational field little g is constant
then you could just use our familiar formula
mgh to find the gravitational potential energy.
But if you're in a region
where the gravitational field is varying
then you have to use this more general formula
which states that the gravitational potential energy
between two masses, m one and m two,
is gonna equal negative of the gravitational constant,
big G, times the product of the two masses,
divided by the center to center distance
between the two masses.
Note, this is center to center, not surface to surface
and it's not squared like it is in the force formula.
This one's just the distance.
Gravitational potential energy
is not a vector but because of this negative sign,
the gravitational potential energy
is always gonna be negative or zero.
It'll only be zero when these spheres become
infinitely far apart
because then you'd divide by infinity
and one over infinity would be zero.
Otherwise, it's always negative.
But even though this gravitational
potential energy is negative,
this energy could still get converted into kinetic energy,
it's just that in order for this gravitational
potential energy to decrease,
it would have to become even more negative
to convert that energy into kinetic energy.
I'm bringing up this topic in this section
because oftentimes when planets are orbiting each other
in circular orbits,
you have to use this formula to determine
the gravitational potential energy between them.
And since it's an energy, the units are joules,
so what's an example problem
involving this more general formula
for the gravitational potential energy look like?
Let's say two spheres of radius R and mass M
are falling toward each other
due to their gravitational attraction.
If the surface to surface distance
between them starts off as four R and ends up as two R,
how much kinetic energy would be gained by this system?
We'll include both masses in our system
and that would mean
that there's gonna be no external work done
so the energy of this system is gonna be conserved.
The system's gonna start off
with gravitational potential energy
negative big G both masses multiplied together
which is M squared divided by the distance
they start off from each other
which is not four R, it's the center to center distance
which is gonna six R.
We'll assume they start from rest
so there'll be no kinetic energy to start with
and this is gonna equal the final
gravitational potential energy negative big G,
both masses multiplied M squared
divided by the distance they end up which is not two R,
it's center to center distance so that's four R,
plus however much potential energy
was converted into kinetic energy.
If we solve this for kinetic energy
we're gonna get negative big G, M squared over six R
plus big G, M squared over four R.
1/4 minus 1/6 is gonna be 1/12.
The amount of potential energy
that was converted into kinetic energy
would have been big G, M squared over 12 R.