- 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
Rotational kinetic energy
David explains what rotational kinetic energy is and how to calculate it. Created by David SantoPietro.
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- 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)
- 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?(4 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(6 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)
- 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)
- [Below is my reasoning and assumption, take with a grain of salt]
If the wheel is only spinning but not rolling (imagine it being held above the ground by its axis), the point at the very center will not have kinetic energy because radius is 0 (I=mr^2, I will equal 0, KE=1/2Iw^2, KE will equal 0). If we set the wheel on the ground and let it roll, then the center will have linear kinetic energy because it is moving forward down the street (displacement occurs). Still, rotational KE at the center will be 0 even though linear KE exists because the center point has a radius 0 from the axis of rotation.(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)
- [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.