# Rotational version of Newton’s second law | Physics | Khan Academy

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– [Instructor] Alright,

so we know how to find the torque now, but who cares? What good is torque? What good is it gonna do for us? Well here’s what it can do. We know from Newton’s second law that the acceleration is

proportional to the force. What we would like to have is some sort of rotational analog of this formula. Something that would tell us alright, we’ll get a certain amount

of angular acceleration for a certain amount of torque. And you could probably guess that this angular

acceleration’s gonna have probably something with

torque on top ’cause torque is gonna cause something

to angularly accelerate. And then on the bottom, maybe

it’s mass, maybe it isn’t. That’s what we need here. If we had this formula,

this rotational analog of Newton’s second law,

then by knowing the torque we could figure out what

the angular acceleration is just like up here by knowing force, we could tell what the

regular acceleration is. So that’s what I want to do in this video. I want to derive this rotational analog of Newton’s second law for an object that’s rotating in a

circle like this cue ball. And not just rotating in a circle. Something that’s angularly accelerating. So it would be speeding up in its rotation or it’d be slowing down in its rotation. So let’s do this, let’s

derive this formula so that if we know the

torque we could determine the angular acceleration

just like we determine regular acceleration by knowing the force and Newton’s second law. So how do we do this? In order to have an angular acceleration we’re gonna need a force that’s

tangential to the circle. So in order to go angularly

accelerate something you need a force that’s tangential because this force is

gonna cause a torque. So let’s say this is the

force causing the torque, we know how to find it now. Remember torque is R

times F times sine theta, but let’s make it simple. Let’s say the angle’s

90 so that sine theta will end up being one

’cause sine of 90 is one. And let’s make it simple too in this way, let’s say this force is the net force. Let’s say there’s only

one force on this object, and it’s this force here. Well we know that the net force has to be equal to the mass of the object times the acceleration of the object. And you’re probably like, big whoop. We already knew this. What’s new here? Well remember, we want to relate torque to the angular acceleration, so let’s write down the torque formula. How do you find the torque from a force? Remember that the torque from a force is gonna be equal to the

force exerting that torque times R, the distance from the axis to the point where the force is applied. Now in this case, that’s the entire radius ’cause we applied this force

all the way at the edge. If this force was

applied inward somewhere, it would be only that

distance from the axis to the point where the force is. But we applied it at the very edge so this would F times the entire radius. And then there’s also a sine

of the angle between F and R, but the angle between F

and R is 90 degrees here, and the sine of 90 degrees is just one, so we can get rid of that. So this is simple, the torque

exerted by this force F is gonna be F times R. What do we do with this? Well look at down here, we’ve

already got an F down here. If you’re creative you might be like, well let’s just multiply

both sides by R down here. That way we’ll get

torque into this formula. In other words, if I

multiply the left side by R I’ll get R times F, and

now that’s gonna equal R times the right-hand side. So it’s gonna be R times

M times the acceleration. And this was good, look

at now we have R times F. That’s just the torque. Torque is R times F, or F times R. So I’ve got torque equals R times M, times the acceleration,

but that’s no good. Remember over here we want a formula that relates torque to

angular acceleration, not a formula that relates

torque to regular acceleration. So what could I replace

regular acceleration with in order to get angular acceleration? Maybe you remember when we talked about angular motion variables. The tangential acceleration

is always gonna equal the distance from the axis to that object that’s got the tangential acceleration, multiplied by the angular

acceleration alpha. So this is the relationship between alpha and the tangential acceleration. Is this tangential acceleration? It is ’cause this was

the tangential force. So since we took the tangential force, that’s gonna be proportional

to the tangential acceleration. These are both tangential here, and these forces are all tangential. That means I can rewrite

the tangential acceleration as R times alpha, and

that’s what I’m gonna do. I’m gonna rewrite this

side as R times alpha ’cause R alpha is the

tangential acceleration. So this whole term right here was just tangential acceleration, and now look what we’ve got. We’ve got torque is gonna be equal to R times M, times R times alpha. I can combine the two

Rs and just write this as M times R squared times

alpha, the angular acceleration. And now we’re close. If I wanted a form of Newton’s second law I could leave it like

this or I could put it in this form over here

and just solve for alpha, and get the alpha. The angular acceleration of this mass is gonna equal the torque

exerted on that mass divided by this weird term, this M the mass, times R squared. And this is what we were looking for. This is what we were

looking for over here. I’m gonna write it in this box. The rotational analog of

Newton’s second law for rotation is this torque divided by this term here. This M R squared, what is that? Well it’s serving the same role that mass did for regular acceleration and the regular Newton’s second law. And remember, this mass was proportional to the inertia of an object. It told you how hard it was to

get that object accelerating. How sluggish an object is. How resistive it is to being accelerated. That’s what this term

down here’s gonna be. People usually call this

the moment of inertia, but that’s gotta be the

most complicated name for any physics idea I’ve ever heard of. I don’t even know what this means. Moment of inertia. That just sounds strange. It’s represented with a letter I, and it’s serving the same role. It’s in this denominator

just like mass is, and it’s serving the same role. It’s serving as the rotational inertia of the system in question. So in other words, something

with a big rotational inertia is gonna be sluggish to

angular acceleration, just like something with

a big regular inertia is sluggish to regular acceleration. So if this ball, and we

can see what it depends on. Look at, for a ball on

the end of a string, the moment of inertia for a

ball on the end of the string was just M R squared. This was the denominator. This was the term serving

as the rotational inertia for this mass on a string. And what that means is

if you had a bigger mass, or if the radius were bigger, this object would be harder

to angularly accelerate. So it would be difficult

to get this thing going and start speeding it up. But on the other hand,

if the mass were small, or the radius were small, it’d be much easier to

angularly accelerate. You could whip it around like crazy. But if the mass were very

big or the radius were big, this moment of inertia

term would get much bigger. This is the moment of inertia for a mass on the end of a string, and

that’s what the I is here. So you could think about it

as the rotational inertia. That’s a much better name for it. People are coming around and realizing that you should just call it this ’cause that’s what it really is. This moment of inertia is

kind of a historical term. It stuck around, it’s not a very good one. Rotational inertia is

much more descriptive of what this I really is. And we should note the units

of this moment of inertia, since it’s mass times radius squared, the units are gonna be

kilgram meters squared. These are the units of moment of inertia, and this is the formula if

you just have a point mass. And by that I just mean a

mass where all of the mass is traveling at the

same radius in a circle. It doesn’t have to be tied to a string. This could be the moon

going around the Earth. But as long as all of the

mass is at the same radius and traveling around in a circle, or at least mostly at the same radius. Let’s assume this little

radius of the sphere is really small compared to

this radius of the string. If that’s the case, where

basically all the mass is traveling around in a

circle at the same radius, this would be the formula to

find the moment of inertia. So how does this ever get harder? What do you have to look out for? Well we only considered one force. You could imagine maybe there’s

many forces on this object. Maybe there’s some other force this way. Well in that case, you just

have the net force here to make sure it’s M times A, and you just have to make sure you use the net torque up here. So this formula will still work

if you have multiple torques on this object or this system. You just have to use

the net torque up here. You add up all of the

torques where torque’s trying to rotate it one

way would be positive, and torque trying to rotate

it the other direction would be negative, so

you’d have to make sure signs are correct up here. And what about rotational inertia? What if your object isn’t

as simple as a single mass? What do you do then? Let’s look at that. Let’s take this formula

here, I’m gonna copy that. Let’s get rid of all of this, and let’s say you had this crazy problem. You had three masses now, and one force on this mass two was 20 newtons downward, and one force was upward 50

newtons on this mass one. And they’re all separated by

three meters, and can rotate. We’re stepping it up, this is complicated. It can rotate in a

circle, but we can do it. We can do it with the

formula we just derived. Let’s use that. This is gonna be useful. Let’s say the question is

what’s the angular acceleration for these masses in this

particular set up of forces? We’re gonna use this formula

for Newton’s second law. In angular form we’ll say

that the angular acceleration if that’s what we want, is

gonna equal the net torque. How do we find the net torque? Now there’s two forces. Well it’s not that bad. You just find the torque

from each one individually and you add ’em up. Just like you would do

with any net vector, find each individually and add ’em up. But it’s not gonna be 50 minus 20. These are torques. We’ve gotta plug torque

in up here, not force. This has gotta be a torque, and until you multiply that

force by an R it’s just a force. So don’t try to just

stick this 50 up in here. It needs to get multiplied by an R. What R? Be careful, you might

think three meters, but no. The R is always from the axis of rotation which is the center all the way to where the force is applied. So the torque from this

50 is gonna be nine meters times the 50 newtons. Now we’ve got a torque. It’s not a torque until you

multiply that force by an R. That was the torque from the 50 newtons. How about the torque from the 20 newtons? You might be like, alright I got it now. It’s gonna be 20 newtons, but

I can’t just put 20, right? We gotta multiply it by an R. It’s gonna be 20 newtons

times, and it’s not three. It’s always distance from the

axis, so it’s from the center all the way to where this

20 newtons was applied, and that’s gonna be six meters. And sometimes when the people

get here they’re just so happy they remember the R, they just do plus, and without thinking about it, but they’re gonna get it wrong. You can’t do that. Look at, this 50 newtons

was trying to rotate this system counterclockwise, right? The 50 newton’s trying

to rotate it this way. The 20 newton is trying

to rotate it that way. They’re opposing each other. These are opposite signs of torque, so I’ve gotta make sure

I represent that up here. I’m gonna represent this 20 newton torque as a negative torque, and that’s the convention we usually pick. Counterclockwise is usually positive, and clockwise is usually negative, but no matter what convention you pick, they’ve gotta have different signs in here so be careful there. So that’s our net torque up here. How do we find the rotational and inertia, or the moment of inertia? Well we know from the previous example the moment of inertia of a point mass that is a mass going in a circle where all of the mass is going

at that particular radius is just M R squared. But now we’ve got three masses so you might think this is

hard, but it’s not that hard. All we have to do is say that

the total moment of inertia is gonna be the sum of all the individual moments of inertia. So we just add up all the

individual moments of inertia. In other words, this is just gonna be the moment of inertia of mass one. If that’s one kilogram,

that’s gonna be one kilogram times R squared. That’s what this means. You take all the masses. M one, times R one squared, plus M two, times R two squared, plus M

three, times R three squared. You’d keep going if you had more. You just add them all up

and that would give you the total moment of inertia

for a system of masses. So if we do ’em one at a time, this one kilogram times the R for that one would be nine meters

’cause that’s distance from the axis to the mass. That’d be nine meters squared

plus alright, mass two. If that’s two kilograms, and that’s gonna be times six squared. And now we keep going. We take this three kilogram

mass and we also add its contribution to the moment of inertia, or the rotational inertia, and that’d be three kilograms times it’s only three meters

from the axis squared, so times three meters squared. And if we add all this up and plug all this into the calculator, we’ll get that the alpha,

the angular acceleration is gonna be 1.83 radians

per second squared. So that’s the rate at which this object would start accelerating

if it started from rest. It would start to speed

up in this direction and start speeding up

over and over and over if these forces maintained the torque that they were exerting. So recapping, just like

Newton’s second law relates forces to acceleration, this angular version

of Newton’s second law relates torques to angular acceleration. And on the bottom of this

denominator isn’t the mass, it’s the rotational inertia

that tells you how difficult it’s going to be to angularly

accelerate an object. And you can find the moment

of inertia of a point mass as M R squared, and you could

find the moment of inertia of a collection of point

masses by adding up all the contributions

from each individual mass.

LorenzoB16Post author6:09

Ali VeliPost authorthank you, nice explanation, i understood

Jimmy ChungPost authorThanks a lot!

Abdul Hakim NorazmanPost authorThanks for this!

Philip YPost authorthis video is by far the best I've seen on this subject!!.. Thank YOU!!.. I wondered how the Radius, R, was used as R^2 in the I= mR^2.. and YOU showed me!!!.. Great video!!..

looksintolasersPost authorThis needs to be retagged! Searching for "Moment of inertia" and "rotational inertia" showed nothing! 70 views! That's criminal!

nIMrOD888Post authorThankyou!

odiupickuPost authorCan rotational system be understood as an inertial given the constant change of the direction of motion?

Ryan LangPost authorI have been stuck on physics for a week and this video made it just click. Thank you so much.