- [Voiceover] When a major league baseball player throws
a fast ball, that ball's definitely got kinetic energy.
We know that cause if you get in the way,
it could do work on you, that's gonna hurt.
You gotta watch out.
But here's my question: does the fact that most pitches,
unless you're throwing a knuckle ball,
does the fact that most pitches head toward home plate
with the baseball spinning mean that that ball
has extra kinetic energy?
Well it does, and how do we figure that out,
that's the goal for this video.
How do we determine what the rotational
kinetic energy is of an object?
Well if I was coming at this for the first time,
my first guest I'd say okay,
I'd say I know what regular kinetic energy looks like.
The formula for regular kinetic energy is
just one half m v squared.
So let's say alright, I want rotational kinetic energy.
Let me just call that k rotational
and what is that gonna be?
Well I know for objects that are rotating,
the rotational equivalent of mass is moment of inertia.
So I might guess alright instead of mass,
I'd have moment of inertia cause in Newton's second law
for rotation I know that instead of mass there's
moment of inertia so maybe I replace that.
And instead of speed squared, maybe since I have
something rotating I'd have angular speed squared.
It turns out this works.
You can often derive, it's not really a derivation,
you're just kind of guessing educatedly but you could
often get a formula for the rotational analog of some
linear formula by just substituting the rotational analog
for each of the variables, so if I replaced mass with
rotational mass, I get the moment of inertia.
If I replace speed with rotational speed,
I get the angular speed and this is the correct formula.
So in this video we needed to ride this cause that
is not really a derivation, we didn't really
prove this, we just showed that it's plausible.
How do we prove that this is the rotational
kinetic energy of an object that's rotating
like a baseball.
The first thing to recognize is that this rotational
kinetic energy isn't really a new kind of
kinetic energy, it's still just the same old
regular kinetic energy for something that's rotating.
What I mean by that is this.
Imagine this baseball is rotating in a circle.
Every point on the baseball is moving with some speed,
so what I mean by that is this, so this point at the top
here imagine the little piece of leather right here,
it's gonna have some speed forward.
I'm gonna call this mass M one, that little piece
of mass right now and I'll call the speed of it V one.
Similarly, this point on the leather right there,
I'm gonna call that M two, it's gonna be moving down
cause it's a rotating circle, so I'll call that V two
and points closer to the axis are gonna be moving
with smaller speed so this point right here,
we'll call it M three, moving down with a speed V three,
that is not as big as V two or V one.
You can't see that very well,
I'll use a darker green so this M three right here
closer to the axis, axis being right at this point
in the center, closer to the axis so it's speed is smaller
than points that are farther away from this axis,
so you can see this is kinda complicated.
All points on this baseball are gonna be moving with
different speeds so points over here that are really
close to the axis, barely moving at all.
I'll call this M four and it would be
moving at speed V four.
What we mean by the rotational kinetic energy is
really just all the regular kinetic energy these
masses have about the center of mass of the baseball.
So in other words, what we mean by K rotational,
is you just add up all of these energies.
You have one half, this little piece of leather
up here would have some kinetic energy
so you do one half M one, V one squared plus.
And this M two has some kinetic energy,
don't worry that it points downward,
downward doesn't matter for things that aren't vectors,
this V gets squared so kinetic energy's not a vector
so it doesn't matter that one velocity points down
cause this is just the speed and similarly,
you'd add up one half M three, V three squared,
but you might be like this is impossible,
there's infinitely many points in this baseball,
how am I ever going to do this.
Well something magical is about to happen,
this is one of my favorite little derivations,
short and sweet, watch what happens.
K E rotational is really just the sum,
if I add all these up I can write is as a sum
of all the one half M V squares of every point
on this baseball so imagine breaking this baseball
up into very, very small pieces.
Don't do it physically but just think about it mentally,
just visualize considering very small pieces,
particles of this baseball and how fast they're going.
What I'm saying is that if you add all of that up,
you get the total rotational kinetic energy,
this looks impossible to do.
But something magical is about to happen,
here's what we can do.
We can rewrite, see the problem here is V.
All these points have a different speed V,
but we can use a trick, a trick that we love
to use in physics, instead of writing this as V,
we're gonna write V as, so remember that for things
that are rotating, V is just R times omega.
The radius, how far from the axis you are,
times the angular velocity, or the angular speed
gives you the regular speed.
This formula is really handy, so we're gonna replace
V with R omega, and this is gonna give us R omega
and you still have to square it and at this point
you're probably thinking like this is even worse,
what do we do this for.
Well watch, if we add this is up I'll have one half M.
I'm gonna get an R squared and an omega squared,
and the reason this is better is that even though
every point on this baseball has a different speed V,
they all have the same angular speed omega,
that was what was good about these angular quantities
is that they're the same for every point on the baseball
no matter how far away you are from the axis,
and since they're the same for every point I can
bring that out of the summation so I can rewrite
this summation and bring everything that's constant
for all of the masses out of the summation so I can
write this as one half times the summation
of M times R squared and end that quantity,
end that summation and just pull the omega squared out
because it's the same for each term.
I'm basically factoring this out of all of these
terms in the summation, it's like up here,
all of these have a one half.
You could imagine factoring out a one half
and just writing this whole quantity as
one half times M one V one squared plus
M two V two squared and so on.
That's what I'm doing down here for the one half
and for the omega squared, so that's what was good
about replacing V with R omega.
The omega's the same for all of them,
you can bring that out.
You might still be concerned, you might be like,
we're still stuck with the M in here cause you've
got different Ms at different points.
We're stuck with all these R squareds in here,
all these points at the baseball are different Rs,
they're all different points from the axis,
different distances from the axis, we can't bring
those out so now what do we do, well if you're clever
you recognize this term.
This summation term is nothing but the total moment
of inertia of the object.
Remember that the moment of inertia of an object,
we learned previously, is just M R squared,
so the moment of inertia of a point mass
is M R squared and the moment of inertia
of a bunch of point masses is the sum of all
the M R squareds and that's what we've got right here,
this is just the moment of inertia of this baseball
or whatever the object is, it doesn't even have to be
of a particular shape, we're gonna add all
the M R squareds, that's always going to be
the total moment of inertia.
So what we've found is that the K rotational
is equal to one half times this quantity,
which is I, the moment of inertia,
times omega squared and that's the formula
we got up here just by guessing.
But it actually works and this is why it works,
because you always get this quantity down here,
which is one half I omega squared, no matter what
the shape of the object is.
So what this is telling you, what this quantity
gives us is the total rotational kinetic energy
of all the points on that mass about the center
of the mass but here's what it doesn't give you.
This term right here does not include
the translational kinetic energy so the fact that
this baseball was flying through the air does not
get incorporated by this formula.
We didn't take into account the fact that
the baseball was moving through the air,
in other words, we didn't take into account
that the actual center of mass in this baseball
was translating through the air.
But we can do that easily with this formula here.
This is the translational kinetic energy.
Sometimes instead of writing regular kinetic energy,
now that we've got two we should specify this is really
translational kinetic energy.
We've got a formula for translational kinetic energy,
the energy something has due to the fact that the center
of mass of that object is moving and we have a formula
that takes into account the fact that something can have
kinetic energy due to its rotation.
That's this K rotational, so if an object's rotating,
it has rotational kinetic energy.
If an object is translating it has
translational kinetic energy,
i.e. if the center of mass is moving,
and if the object is translating and it's rotating
then it would have those both of these kinetic energies,
both at the same time and this is the beautiful thing.
If an object is translating and rotating and you want
to find the total kinetic energy of the entire thing,
you can just add these two terms up.
If I just take the translational one half M V squared,
and this would then be the velocity of the center of mass.
So you have to be careful.
Let me make some room here, so let me get rid
of all this stuff here.
If you take one half M, times the speed of the center
of mass squared, you'll get the total translational
kinetic energy of the baseball.
And if we add to that the one half I omega squared,
so the omega about the center of mass you'll get
the total kinetic energy, both translational and rotational,
so this is great, we can determine the total kinetic energy
altogether, rotational motion, translational motion,
from just taking these two terms added up.
So what would an example of this be,
let's just get rid of all this.
Let's say this baseball, someone pitched this thing,
and the radar gun shows that this baseball was hurled
through the air at 40 meters per second.
So it's heading toward home plate at 40 meters per second.
The center of mass of this baseball is going
40 meters per second toward home plate.
Let's say it's also, someone really threw the fastball.
This thing's rotating with an angular velocity
of 50 radians per second.
We know the mass of a baseball, I've looked it up.
The mass of a baseball is about 0.145 kilograms
and the radius of the baseball, so a radius of a baseball
is around seven centimeters, so in terms of meters that
would be 0.07 meters, so we can figure out what's
the total kinetic energy, well there's gonna be
a rotational kinetic energy and there's gonna be
a translational kinetic energy.
The translational kinetic energy, gonna be one half
the mass of the baseball times the center of mass speed
of the baseball squared which is gonna give us one half.
The mass of the baseball was 0.145 and the center of mass
speed of the baseball is 40, that's how fast the center
of mass of this baseball is traveling.
If we add all that up we get 116 Jules of regular
translational kinetic energy.
How much rotational kinetic energy is there,
so we're gonna have rotational kinetic energy
due to the fact that the baseball is also rotating.
How much, well we're gonna use one half I omega squared.
I'm gonna have one half, what's the I, well the baseball is
a sphere, if you look up the moment of inertia of a sphere
cause I don't wanna have to do summation of all
the M R squareds, if you do that using calculus,
you get this formula.
That means in an algebra based physics class
you just have to look this up, it's either in your book
in a chart or a table or you could always look it up online.
For a sphere the moment of inertia is two fifths M R squared
in other words two fifths the mass of a baseball
times the raise of the baseball squared.
That's just I, that's the moment of inertia of a sphere.
So we're assuming this baseball is a perfect sphere.
It's got uniform density, that's not completely true.
But it's a pretty good approximation.
Then we multiply by this omega squared,
the angular speed squared.
So what do we get, we're gonna get one half times
two fifths, the mass of a baseball was 0.145.
The radius of the baseball was about, what did we say,
.07 meters so that's .07 meters squared and then finally
we multiply by omega squared and this would make it
50 radians per second and we square
it which adds up to 0.355 Jules
so hardly any of the energy of this baseball
is in its rotation.
Almost all of the energy is in the form of translational
energy, that kinda makes sense.
It's the fact that this baseball is hurling toward
home plate that's gonna make it hurt if it hits you
as opposed to the fact that it was spinning when
it hits you, that doesn't actually cause as much damage
as the fact that this baseball's kinetic energy
is mostly in the form of translational kinetic energy.
But if you wanted the total kinetic energy of the baseball,
you would add both of these terms up.
K total would be the translational kinetic energy
plus the rotational kinetic energy.
That means the total kinetic energy which is the 116 Jules
plus 0.355 Jules which give us
116.355 Jules.
So recapping if an object is both rotating
and translating you can find the translational
kinetic energy using one half M the speed of the
center of mass of that object squared and you can
find the rotational kinetic energy by using
one half I, the moment of inertia.
We'll infer whatever shape it is,
if it's a point mass going in a huge circle
you could use M R squared, if it's a sphere
rotating about its center you could use two fifths
M R squared, cylinders are one half M R squared,
you can look these up in tables to figure out
whatever the I is that you need times the angular
speed squared of the object about that center of mass.
And if you add these two terms up you get the
total kinetic energy of that object.