As a flywheel spins faster does it take more power to accelerate it? Let’s say I accelerate my flywheel from 1000 to 2000 rpm in 10 seconds. Will it take more power to accelerate it from 2000 to 3000 rpm in 10 seconds? As per standard physics questions ignore friction and relativity. Note, I’m a long time over 40 doper so this is not a homework question, just a friendly argument with a coworker. I think it will take the same amount of power.

Sorry, but your coworker is right.

Angular kinetic energy is proportional to the rate of rotation squared. So it will take 5/3 times as much energy to go from 2000 to 3000 rpm as it takes to go from 1000 to 2000 rpm and 5/3 times as much power to accomplish both in the same amount of time.

The same holds true for linear motion too. It takes more energy to accelerate a car from 20mph to 30 mph, than it does to accelerate from 10 mph to 20 mph.

e[sub]k[/sub]=1/2 m v[sup]2[/sup]

Wait what?

Are you two sure about that? It sounds very wrong to me. I’ll admit I have little training in this area.

Take for example the ultimate weightless, frictionless environment. An asteroid in space. I give the asteroid a nudge, and it changes speed by 10mph. Now if I preform the same nudge again, surely it’ll change speed by 10mph again?

Okay, that was mainly aimed at Amblydoper, but shouldn’t something similar apply to something rotating too?

Or are you two talking about an environment with friction?

Think about it this way. If the asteroid is going 10 mph and you nudge it up to 20mph, that means when it left contact with your hand, your hand was moving at 20mph, right…it’s not going to gain any more speed after it leaves your hand. Now, if you push it exactly the same way, it’s not going to go any faster. For it to go faster you’ll have to push it harder. To get it up to 30, you’ll have to nudge it in such a way that when it leaves contact with your hand, your hand is going 30mph.

How about this. Go flip a bike upside down so you can spin the front wheel with your hand. Get it going as fast as you possibly can and you’ll find two things…1)The faster it’s going the harder it is to get it going even faster and 2)If you keep spinning it using the same amount of energy it won’t continue to go faster, it’ll stay at the same speed.

To take your example to it’s logical conclusion if you keep nudging something, be it a flywheel or an asteroid, it should keep speeding up…to an infinite speed.

Well bring the hand with you.

A spaceship in space. You fire the engines for 10 seconds, and go 10mph faster. You fire the engines again for 10 seconds, and this time you only go 9mph faster? It just doesn’t make sense to me.

I’m not saying you’re wrong and I’m not saying you’re right. I’m just saying that if you’re right, I’m not convinced.

So if every time you fire the engine for 10 seconds you go 10mph faster could you get going to a million miles per hour? a billion?

Then how much work is the spaceship doing as opposed to your hand? You’re asking it to help you match your frame of reference, aren’t you? How much work is that?

You’re confusing two concepts: *impulse* and *energy*. If you apply the same amount of force for the same amount of time, that corresponding to giving the object a certain *impulse*, which causes a change in the momentum of the object. Since momentum is proportional to velocity, that means that firing your engines for a fixed amount of time will always change your velocity by a certain amount, regardless of what velocity it started at.

But that’s not what the OP asked. The original question was whether the same amount of *energy* was needed to change the velocity by a certain amount. Energy equals force times distance, rather than force times time. If you fire your rockets while covering the same *distance*, you’ll always give yourself the same amount of extra kinetic energy. The problem is that when you’re moving faster, it takes less time to cover that same fixed distance, and so you end up firing your rockets for less time. Which means that you end up with less *impulse*, and therefore less change in velocity.

N.B.: Everything I’ve mentioned above deals with linear momentum and translational kinetic energy; but the same discussion applies to rotational motion with the following substitutions:[ul][*]For “kinetic energy”, read “rotational kinetic energy”[*]For “velocity”, read “angular velocity”[*]For “impulse”, read “angular impulse”[*]For “momentum”, read “angular momentum”[*]For “force”, read “torque”[*]For “distance”, read “angle”[/ul]

Let’s start with the math, which was mentioned earlier. For linear motion, the kinetic energy of an object is proportional to the square of its speed (for rotation, it’s proportional to the square of RPM).

Suppose a flywheel rotating at 1000 RPM has K amount of kinetic energy. Something rotating at 2000 RPM has 4K amount of kinetic energy as compared to rotating at 1000 RPM (K * 2000^2/1000^2). Something rotating at 3000 RPM has 9K amount of kinetic energy as compared to rotating at 1000 RPM (K*3000^2/1000^2).

OK, so to go from 1000 to 2000 RPM, you need to supply 3K mechanical work (4K- K). And to go from 2000 to 3000 RPM, you need to supply 5K mechanical work (9K - 4K).

Baracus said pretty much the same thing, but I figured I ought to show where the numbers came from.

The asteroid nudge problem:

Asteroid starts at 10 MPH. Nudge it with 10 pounds of force for 10 seconds. Suppose the mass of the asteroid is such that this causes it to accelerate up to 20 MPH. OK, how much mechanical work did you just do on the asteroid? Work equals force (10 pounds) times distance. Distance here is how for the asteroid moved while you were pushing on it. In this case, the average speed (while you were pushing on it) would be 15 MPH (22 feet per second). So you pushed on it for 220 feet. Total mechanical work supplied by your hand, 2200 pound-feet.

Now your asteroid is moving at 20 MPH. Nudge it again, ten pounds, ten seconds. Same force, same mass, so same acceleration for same amount of time, so same change in velocity, so you get it up to 30 MPH. This time while you were pushing, the average speed was 25 MPH; so over this ten-second nudge, you pushed on it for 366.67 feet. Total mechanical work supplied by your hand, 3666.7 pound feet. You can see now that for a given absolute change in velocity, the required energy is greater if your initial velocity is greater.

You can quantify your “nudge” differently if you want: X amount of force for Y amount of time, X amount of force for Z amount of distance. Either way, if you can figure out how hard you were pushing and over what distance, you can figure out how much work you did, and ultimately you’ll find that it meshes with the equation for kinetic energy of a moving object, KE = 0.5 x mass x velocity^2.

[quote=A spaceship in space. You fire the engines for 10 seconds, and go 10mph faster. You fire the engines again for 10 seconds, and this time you only go 9mph faster? It just doesn’t make sense to me.[/quote]

No, every ten-second burn will give you an additional ten MPH of forward speed, just as surely as your hand did in the previous scenario. Same as with the hand though, if the initial velocity is higher, the force of the rocket engine is applied over a greater distance (doing more mechanical work on the spacecraft). I know this seems paradoxical, since the rocket is burning the same amount of fuel per unit-time, regardless of the spacecraft’s forward velocity. The key is the absolute velocity of the rocket exhaust. If the spacecraft is stationary, the exhaust is getting shot out the back at, say, 3000 MPH. Lots of kinetic energy going away with that exhaust. If the spacecraft is moving forward at 3000 MPH, then the rocket exhaust is basically left standing motionless once it leaves the engine; all of the kinetic energy that it had a moment ago (when it was aboard the fast-moving spacecraft) gets left on the spacecraft, concentrated among the spacecraft’s now-reduced total mass. So, same amount of energy from combustion, but the spacecraft’s forward speed determines how much of that energy gets pissed away in fast-moving exhaust, and how much of it gets converted to mechanical work in speeding up the spacecraft.

If the spacecraft is moving at 6000 MPH, then the exhaust still has a *forward* velocity of 3000 MPH when it exits the rocket engine. This represents wasted energy: if you could have expelled it from the rocket at 6000 MPH (leaving it motionless), then your spacecraft would have gotten the most energy it could out of it. See propulsive efficiency for a more complete explanation.

I agree with folks that it does. Rocket ships are out of the question; the OP asked about flywheels, which are only necessary in a gravity environment. **Joey P’s **bike example is correct.

Step 1:

I manually spin a flywheel from 1000 to 2000 rpm.

Step 2:

I reset my body to a zero-motion starting position. (this is where the rocket ship example diverges I believe)

Step 3:

I manually spin the fly wheel from 2000 to 3000 rpm

To make Step 3 work, I have to manually accelerate my hand to a faster velocity just to catch up to where the fly wheel is currently spinning before I can even impart a higher rpm. Since the flywheel in step 3 is spinning faster than in Step 1, I have to use more energy to catch up(which will translate into power). Case closed.

Knowing that most folks have a real confused understanding of the technical meaning of the terms “energy”, “power”, “force”, “work”, and “momentum”, I’m going to bet that when the OP & his coworker get back to discussing this they’ll discover they don’t agree on which question they were really asking.

In fact, the OP speaks only to “power”, whereas all the answers refer to “energy”. Related, but not the same.

To make a flywheel spin faster you apply a *torque*. Torque is the change i angular momentum, and is equal to to moment of inertia, **I* times the angular acceleration, **alpha**. There are lots of ways of applying torque, but you’ve already said that you’re applying Force (**F**) at the edge of your wheel, which we’ll assume has radius **r**. The torque equation is then

**Fr = I(alpha)**

well, **alpha** is just the change in angular rate of rotation (usually written small **omega**) with time, so let’s simply assume it’s linear, and

**alpha = (delta omega) /(delta time)**

Here **delta omega** is the change in the rotation rate, which is what you’re looking for. Inserting this in the first equation and re-arranging gives you

**F r (delta time) = I (delta omega) **

or **(F r (delta time)/I = delta omega **

So the change i the rate of rotation (delta omega) is the product of the applied force times the wheel radius times the time over which you apply that force, all divided by the moment of inertia. If you apply the same force at the same point in the same amount of time, you will increase the rate of rotation by the same amount, no matter how fast it’s already spinning. Which I think is what the OP thought.

“Power” is technically the amount of Work done per unit Time. Work is Force exerted over distance. We’ve already assumed that you apply the same force over the same amount of time, so if it acts over the same distance then applying the same power will result in the same change in rotation. Although it’s possible to make things so that this isn’t the case, I suppose, by making the distance different. No doubt the OP wasn’t thinking about these technical distinctions. If you apply the same force over the same interval of time at the same point on the flywheel, you will change the angular rate of rotation by the same amount.

You won’t change the *energy* by the same amount, but I’m not going to go there.

Flywheels can be necessary in space.

Quite right. You can used flywheels to change orientation in space without having to use propellant. They can also be used for energy storageThere are companies that manufacture such device for use on satellites. You can find the idea in old SF stories, too. Heinlein uses it several times.

Correct, the OP did ask about power, and maybe we’ve muddied things by not providing the specific answer to his question.

He asked whether it takes more power to go from 2000-3000 RPM in ten seconds than it does to go from 1000-2000 RPM in the same amount of time. We’ve shown that it takes more *energy* to do so; power is simply energy delivered per unit time, so given that a greater amount of energy is supplied during the same 10-second time interval, it follows that, yes, a greater amount of power is required.

Another way to look at it:

Accelerating by 1000 rpm WILL take the same torque for the same time, regardless of the starting speed.

However, the wheel will spin many more revolutions during that time if it starts out faster. The work is torque X angle, NOT torque X time. Torque X time is impulse.

regardless, if he applies the same force for the same time the change in revolutions will be the same. It doesn’t matter if the wheel is rotating faster. He said “power”, but I doubt if he had the technical definition in mind.

Fair enough. Statement withdrawn.

I have a problem I’m trying to solve without having to actually build the darn thing. Should be ‘simple’ but my brain just isn’t getting wrapped around the math.

Could someone help me?

Flywheel with 2 of 5 lb weights balanced 180° apart at 37.405942346 inches diameter.

Axle is 2.5816111112 inches diameter

Axle has 46.4 lbs torque delivered continously,

say via a string that is wrapped several times around the axle and has 46.4 lbs hanging on it.

Question:

What will be the fps that the 5 lb weights will be moving when the axle turns 180°

How much kinetic energy will be stored with each 180°?

Thank you.