- Introduction to torque
- Moments (part 2)
- Finding torque for angled forces
- Rotational version of Newton's second law
- More on moment of inertia
- Rotational inertia
- Rotational kinetic energy
- Rolling without slipping problems
- Angular momentum
- Constant angular momentum when no net torque
- Angular momentum of an extended object
- Ball hits rod angular momentum example
- Cross product and torque
In this video David explains more about what moment of inertia means, as well as giving the moments of inertia for commonly shaped objects. Created by David SantoPietro.
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- why is calculus required for everything(21 votes)
- Calculus is the mathematical technique to deal with values that change and since most real world situations are not static you need to use calculus to describe what happens.(73 votes)
- How does all this play with torque, i mean like we know that its easier to rotate something with a large amount of torque and that would be the further you are from the axis of rotation hence increasing the moment arm, but the further you are from the axis the larger the moment of inertia will become. So its kind of like both of these contradict one another , the torque is saying its easiest to push here, while the inertia is telling us its resisting angular acceleration the most here?(10 votes)
- Thats a cool question.
Its probably better for you if you can figure it out your self.... let me give you my answer far below, and take a look after you have given it some more thought...
OK; you got here... :)
So, you are right, the moment or toques WILL increase with distance from the turning point.
You are also right that the moment of inertia DOES depend on distance of the mass from a turning point. BUT... the moment of inertia is a property of the body. It will not change as you change the position of your force.
In much the same way that the mass of a body will not change as you change the direction or magnitude of a force acting on it.
Does that make sense??
How did you do?(8 votes)
- At9:28the formula of the moment of inertia. He said he used calculus
to derive the formula I=1/3ml^2. However, when i tried deriving it using the indefinite integral. I got I= 1/3ml^3. Could you please explain how you got
- I think you probably assumed that I=∫mr^2 dr =∑m_ir_i^2. But look, the masses in the sum are getting smaller and smaller as you increase the number of parts you divide the rod into (eventually reaching infinitely many, infinitely small point masses - our goal). On the other hand, in the integral, each point mass has the mass m. It would imply that the whole rod has infinitely high mass, which makes no sense.
My solution is as following - introduce a linear density σ such that σ=m/L, so m = σL. This way, every infinitely short part of the rod will have an infinitely small mass, but together, their masses will sum up to m. Consequently, I = ∫(from 0 to L) σr^2 dr = 1/3 σL^3 = 1/3 (σL)L^2 = 1/3 mL^2.(6 votes)
- What will be moment of inertia of rod whose axis of rotation is somewhere between the centre and the edge?
In this case, do we have to consider the larger length or the smaller length?(5 votes)
- We are basically rotating two rods with lengths x and L-x.
The moment of inertia would be
mx³/(3L) + m(L-x)³/(3L).
Note: If x = L/2, we get mL³/(24L) + mL³/(24L) = mL²/12 as seen in the video.(2 votes)
- If we consider a door rotating about it's hinge, isn't it more difficult to rotate it near its axis than from its end point? Will that not mean that moment of inertia is greater near its axis ?(2 votes)
- The reason that it takes more force near the hinge is that the torque that you exert is the product of the force (perpendicular to the door) exerted times the torque arm (the distance from the hinge). The same torque requires more force if the distance is less. So the angular acceleration is lower (for the same moment of inertia) when the torque is lower.(5 votes)
- Hi! So at9:00when you wrote the formula for the rod of constant density rotating around the axis at one end, I had expected it to be (mL^2)/4.
I thought that since the center of mass was at the exact center of the body, we could consider the system to be a rope with a body of mass 'm' situated at distance 'L/2'. That way it would be (mL^2)/4.
Could anyone tell me where I went wrong? Ik he said that we were supposed to use calculus, but I can't understand why my explanation isn't correct.(3 votes)
- It's because of L square. If there was just L (not-squared), you would be right. I am sure you will understand this when you learn calculus, but let me give you an example.
Consider just 3 point masses for simplicity. One is at the axis of rotation (like the left end of the rod), the third is at the distance L from the axis, and the second point lies exactly between them.
Each has a mass m, so the total mass is obviously M=3m and m = 1/3 M.
Following your way of thinking, the mean distance from the axis of rotation is L/2 (equal to (0 + L/2 + L)/3), so the moment of inertia would be
I = ML^2/4.
However, let's now consider these points separately. The moment of intertia of the first point is i1 = 0 (as the distance from the axis is 0). Of the second point: i2 = m (L/2)^2 = mL^2/4. Of the third point: i3 = mL^2. The total moment of inertia is just their sum (as we could see in the video): I = i1 + i2 + i3 = 0 + mL^2/4 + mL^2 = 5mL^2/4 = 5ML^2/12.
The result is clearly different, and shows you cannot just consider the mass of an object to be concentrated in one point (like you did when you averaged the distance). You can also check that it would not matter if the distance wasn't squared - then your method would produce correct results.
I hope this clears things up. Cheers(2 votes)
- question-we have a beam , o is axis of rotation f1 is, first force's position direction is upward while f2 is second force's position . whole of beam from f1 to f2 is massless so can we say f1=f2?
- are we saying the beam is in equilibrium??
If so, then no. We can not say f1 - f2.
We can say that the turning effect due to f1 = - (minus) turning effect due to f2
and the turning effect (or moment) of each force is given by = Force x distance from turning point
when we are told a beam is massless, it means the weight of the beam is tiny compared with the other forces so you can neglect it.
- I need to learn the proofs of the differential and integral calculus for these results?
Can anyone provide me with them ?(1 vote)
- If radius is bigger, than according to moment of inertia it is more difficult to rotate but we also know that the bigger the radius the more the torque ........... than the simple question is how radius effects the rotation........and why?(1 vote)
- Moment of inertia and torque are completely independent of each other.
If you spread out the mass distribution further away from the axis of rotation, the moment of inertia will increase. If you apply a force further away from the axis of rotation, the torque will increase. Even though both depend on radius, the torque and moment are completely independent (ex. you can apply the force anywhere or the mass can be distributed anywhere).(2 votes)
- according to the inertia formula it's gonna be harder to open the door from the side(where the door knob is ); because its further than the rotation point (axis), than opening it from the opposite side; because its near to the rotation point.
while i feel it's much easier!!
can you please explain?(1 vote)
- The moment of inertia of the door doesn't change based on where you push it, the center of mass of the door is not changing its distance from the pivot point.
The change in how hard it is to move has to do with leverage not moment of inertia.(2 votes)
- [Instructor] We should talk some more about the moment of inertia, 'cause this is something that people get confused about a lot. So remember, first of all this moment of inertia is really just the rotational inertia. In other words, how much something's going to resist being angularly accelerated, so being sped up in its rotation, or slowed down. So if it has a, if this system has a large moment of inertia, it's going to be very difficult to try to get this thing accelerating, but if the moment of inertia is small, it should be very easy, relatively easy to get this thing angularly accelerating. So that's what this number is good for, the reason why you wanna know the moment of inertia is 'cause it'll let you determine how difficult it'll be to angularly accelerate something, and remember it shows up in the angular version of Newton's second law, that says that the angular acceleration is gonna be equal to the net torque divided by the moment of inertia, or the rotational inertia, since they're the same thing. So that should make sense, we're dividing by the moment of inertia, we're dividing by the rotational inertia because that means if this rotational inertia is big, look it, this is in the denominator. You've got a big denominator, you're gonna have a small value, that means this alpha is gonna be small, it's gonna be a small angular acceleration, but if this moment of inertia were small, then it's gonna be easier to rotate, and you'll get a relatively larger angular acceleration 'cause you're now dividing by a smaller number. So it does serve the same role that mass did, it serves as this inertia term for angular acceleration, and we figured out how to determine the moment of inertia for a point mass, and you'll hear people say this a lot, "point mass," I'm gonna say it a lot. By point mass I just mean a mass you could treat as if all the mass were rotating at the same distance from the axis, and that's what's happening here. If you've got a heavy ball connected to a string, a very light string that has very little mass, you can neglect the mass here. If all the mass is rotating at the same radius like this is, we determined last time that the moment of inertia of a point mass going in a circle is just the mass times how far that mass is from the axis, squared. This is the term for a point mass going in a circle for what the moment of inertia is, how difficult it's going to be to angularly accelerate. This is the rotational inertia, mr squared, but you get more complicated problems too, so you could be like, "All right, what happens "if we don't have a single point mass, we've got the three?" Well we did this last time as well, if you have multiple point masses, all you need to do is say that all right, for multiple point masses, just add up all the contributions from each individual point mass. So if we're careful here, mathematically, we should put an i subscript, but don't let that freak you out, this just really means all them all up. So this would be m one times r one squared, so you take the mass one times its distance from the axis squared, plus m two times r two squared, you take mass two times its distance from the axis squared, and then you do the same for m three, and if you had more masses, you would just keep adding 'em up. If you have a whole bunch of point masses that you can treat as if all the mass were rotating at the same distance from the axis, and you might object, you might say, "Wait, "different masses here are rotating "at different distances from the axis," but all of that particular mass, all of m one is rotating at the same radius from the axis, so we can use this formula for point masses and we can add them up. The total amount is gonna be the total rotational inertia, so in other words, for this case here, if we really wanted to do it, we would say that the moment of inertia for these objects, and this system in total would be, all right, let's take 'em in order. M one is gonna contribute m one times its distance from the axis squared would be a, so we do a squared, and let's say b is just the length of this string, so b just represents that length, and similarly c represents that length, and we're gonna assume the radii of these masses are small. I had to draw 'em big so we could see 'em, but it's easiest if you consider them to be small, 'cause then we don't have to take into account their actual radius. So we'd add to this, that's this m one a squared is just the contribution to the moment of inertia that's being contributed by just m one, so we have to figure out the contributions from each of these other masses, so we'll have m two times its distance from the axis. It isn't gonna be b, it's gonna be all the way, so that's gonna be a plus b squared, and then if you wanted to find the contribution from m three so that you'd get the total, you'd have m three times, well, it'd be a plus b plus c squared, this would be the total moment of inertia for the entire system, which says it's gonna be more difficult, right? The more mass you add into the system, the more sluggish it is to acceleration, the more difficult it is to rotate. So how could we make this three mass system easier to rotate? Let's say you were tired of requiring so much torque to move this thing, you wanna make it easier to rotate. One thing you can always do is just take your masses and move them toward the axis, i.e. just move these toward the center. If you do that notice all of these rs are gonna get smaller, if you reduce the r you're gonna get less moment of inertia, and that object's gonna be easier to rotate, easier to angularly accelerate, you can whip this thing around easier if the mass is more toward the axis. So this makes sense, think about a baseball bat. If you had a baseball bat, so if you got this baseball bat, this is not the best drawing of a baseball bat, but you've got a baseball bat. If you swing it from this end where this is the axis, it's hard to rotate, right? You've got all this heavy mass over here at the end, but if you swing it instead where this is the axis, if you just turn it around and swing it from this end, where this is the axis, now you've made it so most of the mass is near the axis, and if you do that, the radius of that mass is gonna be smaller, and if the radius is smaller it's gonna contribute less to the moment of inertia, less to the rotational inertia, it's gonna be easier to swing. So you can swing a baseball bat really easy if you hold it by the fat end, compared to the actual end you're supposed to hold, you can swing this faster. It's probably not a good idea, you've probably not gonna hit the ball very far, but you'll be able to swing it much faster 'cause that moment of inertia's gonna be smaller. And then the other thing we could do, we could always just reduce the masses. If you can make the mass less you reduce the moment of inertia, and if you can move those masses toward the axis, you reduce the r, you reduce the moment of inertia or the rotational inertia. But what if you don't have point masses at all? I mean, we don't always have situations where the thing that's rotating are a bunch of point masses, what if you had something more like this, where it was like a rod that had its mass evenly distributed throughout the entire rod, and it rotated in a circle. I mean, we couldn't use this formula now because this assumes that all the mass is rotating at some radius, r, but for this rod, only the mass at the end of the rod is rotating at the full length of the rod. The mass that's closer to the axis is gonna have a smaller radius, it'll only be rotating at part of the length. This would only have a radius of L over two, and this part right here would only have a radius of maybe, L over eight. So how do we figure this out? We can't just say the total mass of this rod, if this rod has a total mass m, and a total length L, we cannot say that the moment of inertia of this rod about its end is gonna be mL squared, that's just a lie. This total mass is not rotating all at a radius of length L, only the little piece at the end is rotating with a radius of length L. The rest of this mass is having its contribution to the rotational inertia diminished by the fact that these masses are getting closer and closer to the axis, so what do we do? Well we can't use this, let's get rid of this. That's not possible. The truth is you have to use calculus to derive the formula for these continuous objects, and it's fun. You can do integrals and you can solve for these moments of inertia, that's one of my favorite calculations to do, it's kinda like a puzzle. You can solve for the moments of inertia, but if you don't know calculus, that would just look like witchcraft to you, so I suggest you learn calculus and try it, 'cause it's really fun, but I'm just gonna give you the result. It turns out the moment of inertia for this rod is gonna be, and without knowing the exact answer, we should be able to say, is it gonna be bigger than, less than or equal to mL squared. We should be able to say, it's gotta be less than mL squared, it's not going to be mL squared, it's gonna be less than this because mL squared would be if all of the mass were at the full length of the rod for their radius. Then you would put mL squared. If you could melt this rod down into just a ball, and put that ball at the very far end, you'd be maximizing its rotational inertia, 'cause you'd put all of the mass with the same largest radius r, but some of this mass is in here. Some of this mass is only at L over two, or L over four, or at L over eight. So those little pieces of mass are having their rotational inertia contribution diminished, so we're gonna have less than Ml squared. How much less? Turns out for a rod about its end, it's 1/3 mL squared, and if you do the integral, that's where this 1/3 comes from. So this is for a rod with the axis at the end of the rod. So that's the moment of inertia for a rod rotating about an axis that's at one of the ends of the rod, but what if we move this axis to the center? What if we move the axis here so that this whole rod rotates around a point in its center. Do you think the moment of inertia of this rod that's the same mass and length that it was, we're just rotating it about the center, do you think this moment of inertia is gonna be bigger than, smaller than or equal to what the moment of inertia was for a rod rotated about the end. And the way I would think about it, I'd just ask myself this question, "Is more of "the mass farther away now, or closer to the axis?", 'cause we know if we can decrease these rs, we decrease the moment of inertia, and in this case we did decrease the rs. Think about it, the farthest some piece of mass will be from the axis now is L over two. It's L over two this way, and L over two that way, whereas before, where the axis was at one end, some of the mass was at L away, so that'd be L squared, but now you're only gonna have L over two squared for the farthest some piece of this mass is gonna be, and that's gonna decrease the moment of inertia even more, because more of this mass is closer to the axis when you move it to the center, so it's gonna be less than 1/3 mL squared. Turns out if you do the integral you get 1/12 mL squared, so this is for a rod with the axis at its center. So what's another common geometry? Well if we get rid of that, another case that comes up a lot is a cylinder, or sometimes it's called a disc. So let's say you have a cylinder, a solid cylinder of mass m and it has a radius r, what would this moment of inertia be? Well you can probably tell by now, all right, so it's not gonna be the total mr squared, and it's not gonna be the total mr squared because all of the mass is not rotating at the full radius of the cylinder, right, so it's gonna be less than this. How much less? If you do that integral it turns out that you get 1/2 mr squared, so it turns out the fact that some of these masses are closer to the axis than the full radius of the cylinder, makes it so that the total moment of inertia is 1/2 the total mass of the cylinder times the total radius of the cylinder squared. This is for a cylinder with the axis through the center, so the center, so it's rotating around a point right here, so it's rotating like this around this point here, and that's important to note. It's not enough to just say, "Hey, I gave you a rod, "what's the moment of inertia?", because you've gotta know, "Well where's the axis?" If someone just hands you something and says, "What's the moment of inertia of this?" You can't give them an answer until they've specified where they want you to rotate the object around. If you rotate the rod about its end, it's 1/3 mL squared for the moment of inertia. If you rotate the rod about the center it's 1/12, and again, the reason for that is 'cause by rotating it around different axes, you've made it so some of the mass is at different rs from other axes that you could choose. So this was for a cylinder, also called a disc. Sometimes a sphere comes up, so this is another common example, say you had a sphere, also rotating around an axis, like the earth rotating on its axis, and let's say it also has a mass m and a radius r. Again, because some of this mass is closer to the axis, look it, this mass right here is only rotating in a circle like that, as opposed to at the full radius of the sphere, it's gonna have less than mr squared. How much less? Well for a sphere rotating about an axis that goes through its center, you get that the moment of inertia is 2/5 mr squared, so that was for a sphere rotating about an axis that goes through its center. And at this point you might object, you might say, "Wait a minute, "we had spheres when we had spheres before, "and we did mr squared," but that was for spheres that were rotating where all of their mass was rotating at the same radius. So if you have a sphere, in other words, if you have a sphere and you're gonna rotate this whole sphere around in a circle like this, if that's the case you're talking about then yeah, that total mass is all rotating at the same radius, but here that's not the case. This is a sphere rotating around its center. So if you just have a sphere that spins in place, that's not the same case as this mass that's being whirled around, around some common axis all at the same radius. It's the difference between, this is like the moon rotating around the earth. If you wanna talk about the moment of inertia of the moon rotating about the earth, you could treat the moon as a point mass, and you'd use mr squared, but if you're talking about the earth rotating on its axis, right? Not the earth going around the sun, but the earth rotating on its axis, then you'd have to say that the moment of inertia for that amount of rotation is 2/5 mr squared, because it's a sphere rotating through an axis that goes through its center. All right, so recapping, the moment of inertia or the rotational inertia gives you a number that tells you how difficult it'll be to angularly accelerate an object. If you've just got a point mass where all the mass rotates at the same radius, you could use mr squared. If you've got a collection of point masses, you can just add up all the mr squareds. If you've got a rod rotating about its end, you could use 1/3 mL squared. A rod rotating about its center is 1/12 mL squared. A cylinder rotating about its center is 1/2 mr squared, and a sphere rotating with an axis through its center is 2/5 mr squared. The reason why all these shapes that have mass distributed through them have factors that make their moment of inertia less than mr squared or mL squared is because some of that mass for a distributed object has mass closer to the axis than a case where all the mass is at the end. So the fact that you've got some of these masses that are closer to the axis for a uniform object reduces the total moment of inertia since it reduces the r, and if you ever forget any of these formulas, there's often a chart in your textbook, or look up the chart online, they're all over the place, lists of all the moments of inertia of commonly-shaped objects, and the axis. You gotta check that it's the axis that you're concerned with as well.