Physics question

If you roll a small rod down the side of a parabola, it will roll to the bottom and up a little on the other side. If you add weighted “flywheels” to each side of the rod, and roll it down again from the same spot, will it go up higher on the other side? If so, why can’t the weighted “flywheels” be made heavy and large enough to eventually make it go back up to the height it was dropped?

You can’t extract more energy that you put in…the flywheels will use energy too, they won’t add in any to the ‘system’ you set up.

I wouldn’t be surprised if you are actually shortening the distance it will travel by adding flywheels that will absorb some energy.

How high the object will roll back up the other side depends entirely on frictional losses, not mass. In a perfectly frictionless system, the object will roll up to the exact height it started from. In the real world, this cannot happen.

Without any friction, the rod would roll back up to exactly the height it was dropped from (and keep rolling back and forth forever). But of course there is friction, so it will roll up less and less each time. Nothing you can do can make it roll higher than it started.

Now if you want to get complicated, it might be possible to add a flywheels and make the rod roll up higher than it does without them, though still lower than it started. If you increase the weight of the rod without increasing friction much, then the relative effect of friction will be less, and so the rod will roll up closer to the starting height.

With zero friction wouldn’t the rod actually slide as opposed to roll?

Brett, the short of the story is that under no circumctances are you going to be able to make the rod roll HIGHER than it’s starting height. You’re up against a law of physics, and you are getting into ‘perpetual motion’ territory…or ‘free energy’ territory, both of which are quack science.

So, while you might be able to get the rod to slide back to it’s original height in a frictionless enviro, not matter what the enviro or what the contraption, you won’t be able to get it higher. And in our world of friction, it will always be lower. Always.

If you read the OP carefully, he never suggests it would be possible for the rod to roll up the other side higher than it started from. I suspect he knows this to be impossible.

Q, don’t discourage someone - me - from posting information which might be helpful. He does inquire, “will it go up higher on the other side?”.

Yes, but clearly he meant “higher than it did without the flywheels”, given the context of the rest of his post.

In the context of his post… he also inquires about getting it back to the same height, which is an indication that he doesn’t grasp a simple concept of physics, so nothing is ruled out, hence my post.

If you make the flywheel at the back heavier than the one at the front, the rod may well end up higher than it started off. However, this does not break any laws of physics.

The incredible Honda cog commercial (warning - Flash animation starts automatically) uses weighted flywheels to move up a ramp. But since outside energy was used to place the flywheels in that position in the first place, no laws of physics are broken.

Discourage? I don’t think Q.E.D. was discouraging you from posting–surely, Q was merely encouraging you to read more carefully.

I’m not 100% sure how friction works, so someone can correct me if I’m wrong. But as I understand it, by adding flywheels, you reduce the height of the rod on the first return. Here’s why.

The frictional force is directly proportional to the normal force, which is proportional to the mass. Since the force is proportional to the mass, the frictional acceleration[sup]1[/sup] is constant (with respect to mass). That is, at a given point on the parabola, two rods of different masses would have the same acceleration. So they’d behave exactly the same.

Now, that’s for two rods with the same dimensions, but when you put flywheels on, or change the size of the rod, you’re changing the moment of inertia[sup]2[/sup]. The fraction of kinetic energy which is trapped in rotation depends on this moment of inertia. The flywheel-laden rod has a greater moment of inertia[sup]3[/sup], so a greater fraction of its kinetic energy will be bound in rotation, so it will be rolling more slowly down the incline. However, the frictional acceleration is no different. The only difference is that it’s experiencing this acceleration for a greater amount of time. Thus it is slowed more by friction per trip, and thus it doesn’t go as high.

Footnotes:
*1 - I realize that the rod only has one acceleration, caused by the sum of the gravitational, normal, and frictional forces. However, it’s not difficult to conceive of an “effective” acceleration which would be due to the frictional force alone.

2 - By “moment of inertia”, in this case, I mean the moment of inertia with respect to the spin axis, not with respect to an axis through the center of mass, which is what you usually use.

3 - Strictly speaking, it has a greater moment of inertia to mass ratio, which is what’s important. But for a given mass, it’s also correct that it has a greater moment of inertia.*

What is the mechanism for getting the flywheels to store some of the kinetic energy? If they’re essentialy wheels, then what you’ll find is the heavier mass will probably reduce the overall friction and the rod will go higher on the other side than it would w/o the wheels. But it all depends on the materials used, etc. I’d need some more details about what you’re trying to do to answer the question.

:smack: Already I see that Quercus has posted something that contradicts me.

Care to explain how this works?

"it might be possible to add a flywheels… "

“Care to explain how this works?”

If you use an unsymmetrically balanced flywheel, you could position the apparatus so that the center of gravity of the rod/flywheel is located higher than the axis of the rod. This provides you with a little extra initial stored potential energy, so the ending state could conveivably have the axis of the rod at a higher point depending upon the orientation of the flywheel (the COG at the end would be located below the rod instead of above).

What I figure is this, but I am by no means certain. First of all let me clearify my original discription a little. When I suggested adding the flywheels to the rod and rolling it, I assumed the device would still be rolling on the rod (same diameter) with the flywheels hanging off the sides of the parabola. This way the rotation of the flywheels would be faster because of the small axis they’re rolling on. Now then, what I think is that there must be some formula that states something to the effect of, by doubling the flywheel size and/or weight, the device will roll half the distance higher then it previously did (halfway between where it last went to and the original dropped height). Therefore, it WILL in fact continue to go higher on the other side with the greater mass attatched to it, however, in smaller and smaller increments. Never to actually get to the ponit of original dropped height.

So does anybody out there have the “Straight Dope” on this matter? Was I somewhat correct on my theory? Will it continue to roll higher with added weight, but in smaller and smaller increments?

No worries.

Instead of flywheels, consider two metal cylinders having the same external dimensions and surface finishes.

One cylinder is hollow, made of aluminium, and has a mass of a few grams.

The other cylinder is solid, made of iridium, and has a mass of several kilograms.

The air resistance (frictional force) that each cylinder encounters is independent of mass. But that force will have a much greater effect on the aluminium cylinder because it is larger in comparison to the gravitational force.