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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1 > Unit 7

Lesson 5: Rotational kinetic energy# Rotational kinetic energy

David explains what rotational kinetic energy is and how to calculate it. Created by David SantoPietro.

## Want to join the conversation?

- How come the units aren't equal on both sides of the formula?

If you write out the units of K = 1/2 I 𝝎^2 in SI-units you get:

kg m^2 s^-2 = kg m^2 rad^2 s^-2

This would mean the rad unit just appears and dissapears. Is this correct? How can this be explained?(12 votes)- Since no one solved your question...

Radians can disappear or reappear in an equation because the unit itself is a ratio (if I remember correctly it should be the ratio of the radius to the arc length where it is the same length as the radius [1 rad is like the 2π in 2πr=C where this time it's just 1r=part of C])].

Due to it being an equal (1:1) ratio, it effectively cancels itself out if you want it to. You can google dimensionless units for more information.

I hope this will help you and any future onlookers of this page!(23 votes)

- At6:38, why we got different masses for different points? Aren't they the same? because they are all on the same ball.(7 votes)
- With calculus, to get the real rotational kinectic energy, you would use all points of the baseball, each with the same infinitely small mass (considering the density is constant for the ball).(2 votes)

- David says that the moment of inertia is mr^2, but isn't the moment of inertia of a sphere 2/5mr^2? Does the summation result in that quantity? Or is 2/5mr^2 only used in special circumstances?(3 votes)
- the moment of inertia of a single point mass a distance r from the turning point is I = mr^2

for all other shapes is more complicated; the half circle is just one example(7 votes)

- In the baseball problem shown in this video, why did the relationship v=ωr fail?

v=40m/s but ωr=50*0.07 m/s=3.5 m/s ! Did I make any mistake here?(3 votes)- In the formula v=wr, v is the tangential speed of a particular point on the baseball.

What you have calculated using the formula is the tangential speed of a point on the outermost edge of the ball for which the w is 50s^(-2) and r=0.07m.

What v=40m/s here in the question given is the translational velocity of the center of mass of the baseball(5 votes)

- If the total kinetic energy is the sum of both kinds of energies, does that mean that the energy I've used to throw the ball is going to be divided between translational and rotational kinetic energy? In an ideal world, if I could minimize the rotation when throwing, would the ball go faster/land further away from me?(2 votes)
- If you could minimize rotation without otherwise reducing the energy you put into the throw, then sure, you would be able to throw the ball harder.

Whether it would go further is a different questions. The spin of a ball causes interesting effects with air resistance and in some cases can cause a spinning ball to pass more easily through the air than a non spinning ball. Golf balls have dimples to maximize the effect of spin on the distance of the golf shot.(7 votes)

- How would you find the moment of inertia of a sphere using calculus? Would you use integration or derivation? And how?(3 votes)
- You would use integration because you are adding up a bunch of little inertias to get the total moment of inertia.(3 votes)

- How could be the angular velocities the same for every point ?(2 votes)
- I think we need to review the basic concepts of rotation here. Each point on a rotating rigid object possesses the same angular velocity but different linear velocities. If the angular velocities weren't the same at every point, the object would break apart.(3 votes)

- So what is the kinectic energy of the speed of the center of a wheel? is it still 1/2 mv^2 ?(1 vote)
- How does changing the radius of an object change its rotational kinetic energy? Logically, it seems like it would increase it because KEr= (1/2)(mr^2)(angular velocity)^2 but wouldn't the radius affect the angular velocity?(1 vote)
- Looking at the equation:

Kr = (1/2) * m * r^2 * ω^2

Without the summation portion this is the rotational kinetic energy of a small piece of the object.

The term r is not the radius of the whole object. It is the distance of the small part of the object we are looking at is away from the axis of rotation.

It is the same with the term m, it is not the mass of the whole object, it is just the mass of the small part we are looking at.

The term ω while being for the small piece we are looking at the whole object has the same angular velocity since it is a rigid object, a point 1/2 the way to the center of the object completes a rotation in the same time that a point on the surface does.

Lets assume that the size and mass of the pieces of the object we are looking at are all the same so mass is a constant.

What happens to the value of Kr for parts of the object at different distances from the axis of rotation (different values of r). For a given ω as the value of r increases the overall value of Kr increases since as r increases the part has to travel around a larger and larger circles.

Now what happens to the value of Kr if we look at different values of ω for the same value of r. As ω increases Kr increases because the piece at distance r is moving around the circle faster.

The angular velocity ω is not tied to the distance from the axis of rotation r unless you want to keep Kr constant.(2 votes)

- Why is the Inertial Mass multiplied by 2/5?(1 vote)
- that's the moment of inertia for a solid sphere of mass m and radius r: I =2/5*m*r^2(2 votes)

## Video transcript

- [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.