If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Rotational version of Newton's second law

CHA‑4.D (EU)
CHA‑4.D.1 (EK)
CHA‑4.D.1.1 (LO)
INT‑3.F (EU)
INT‑3.F.2 (EK)
INT‑3.F.2.1 (LO)
David explains moment of inertia and the rotational version of Newton's second law and shows how to solve an example problem. Created by David SantoPietro.

Want to join the conversation?

  • piceratops seedling style avatar for user bobby glinton
    should we not find the torque of object #3?
    (14 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user danyoon1019
    I got (alpha) = 2.167m/s^2 for the last problem. What did I miss and how would I make the answer into rad/s^2?
    (6 votes)
    Default Khan Academy avatar avatar for user
    • leafers tree style avatar for user majka.zelenayova
      upper line: you end with 450-120 (N.m)=330 N.m
      Newton: I think about it from the basic formula F = mass times acceleration (F=ma). From this you take just the units: Newton = kg . m/s^2, so now you know how to write Newton in basic units.
      bottom line - unit of the Moment of inertia: I=m.r^2, so unit is kg.m^2

      So up you have: N.m = kg . m/s^2 (N) times meter = kg. m^2/s^2
      down you have: kg.m^2

      Kilograms will cancel out, also meters square (m^2) will cancel out. What will last is 1/s^2.

      But I would like know too, how we got radians. I guess it is because we callculated how big angle have masses come by the speed that they have earned from the applied forces.
      (13 votes)
  • starky ultimate style avatar for user Alex H
    When he derives the rotational acceleration formula, isn't he assuming that the angle is 90 degrees and that sin theta is 1? What if it isn't 90 degrees? Would the formula be torque/mr^2sin(theta)?
    (5 votes)
    Default Khan Academy avatar avatar for user
  • starky ultimate style avatar for user Michele Franzoni
    I'm probably going to embarrass myself with this question but still, here it comes.

    If the rotational inertia is proportional to the square of r (aka the distance vector) how come that it's so much easier to open a door pushing from the knob? If i make the door swing exerting a force on a point closer to the "lateral" edge (thus increasing r) it requires much less effort than, say, pushing/pulling really close to the axis around which the door is rotating.
    Is common sense failing me or am i missing something important here?
    (9 votes)
    Default Khan Academy avatar avatar for user
    • starky seed style avatar for user Dishita
      Shorter version:
      Torque is not equal to F but dependent on it.
      Constant T -> greater the radius->lesser the ANGULAR acceleration
      Basically, rate of change of the rate of change of the angle is lesser.
      You can try pushing a door with the same T, the closer you are, the greater the angular displacement, extend that idea to angular acceleration.
      Good question XD
      correct me if wrong
      (1 vote)
  • male robot donald style avatar for user Tilak Madichetti
    How to know if it is rad or deg ?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • mr pink red style avatar for user Saihaj
    Why is it that when you are further from the axis, angular acceleration increases? Looking at the formula for angular acceleration, if r increases, wouldn't rotational inertia increase even more than torque because rotational inertia is mr^2 while torque is just Fr? I'm asking because for the practice called "Angular Acceleration and angular second law" it repeatedly says that increasing the radius increases angular acceleration because it increases torque.
    (6 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Tyler Jeffries
    How come he didn't take gravity into account.

    Doesn't the first mass (3) have a force of 30 Newtons? Because 3kg * 10 m/s/s = 30 Newtons.

    Therefore you could calculate torque. 3 meters * 30 newtons = 90 NM

    And then you could do the rest for all three of the masses, and also get different torques for the other two while taking gravity into account.
    (2 votes)
    Default Khan Academy avatar avatar for user
    • primosaur ultimate style avatar for user Leafarian
      He is not taking gravity into account. He is just saying there are only two forces acting on two of the balls. This is in an alternate dimension and will probably never happen. He is trying to simply things. In real life, there would be so much to take into account it would be nearly impossible to get the correct answer!
      (6 votes)
  • leaf green style avatar for user david romine
    What effect does the centripetal acceleration have on the angular acceleration? As the mass begins pulling on the string and the string on the mass does that deduct from the total angular acceleration? Is the centripetal force accounted for in this equation?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user Mark Zwald
      Short answer is they are related through angular velocity, but torque causing angular acceleration and rotation causing centripetal acceleration are independent of each other. The amount of centripetal acceleration does not impact the amount of angular acceleration.
      The angular acceleration is α(t) = dω/dt.
      The centripetal acceleration is a(t) = v²/r = ω²r.
      Rearranging terms
      ω = √(a/r)
      α(t) = dω/dt = d/dt[√(a/r)] = 1/[2r*√(a/r)] * da/dt
      Not a particularly useful result but that is how they are related.
      (3 votes)
  • leaf red style avatar for user I.kar
    Are we assuming the system in the example to be taking place in a horizontal plane? Else mg would come under consideration right?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leaf orange style avatar for user Orangus
    Can't you just get rid of r's?
    I mean a = T/I = Fr/mr^2 = F/mr.

    If not, why?
    (3 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [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.