- [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.