Rotational kinetic energy | Moments, torque, and angular momentum | Physics | Khan Academy

- [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.