- 05 May '08 16:35 / 1 editThis time we have a hollow sphere that weighs 10 pounds.

We pour 2 pound of sand inside the hollow sphere but do not fill up the entire void inside the sphere.

The bridge can only support 11 pounds before it collapses instantaneously.

Using the concept of centrifugal force (ie sand in a bucket) is there a speed that you can roll the sphere such that the sand will spread out around the inside of the sphere and thus reduce the amount of downward force as the sphere travels along the bridge?

Make any assumptions you need to that are realistic. - 05 May '08 17:01 / 1 edit

No, the bridge will not hold.*Originally posted by uzless***This time we have a hollow sphere that weighs 10 pounds.**

We pour 2 pound of sand inside the hollow sphere but do not fill up the entire void inside the sphere.

The bridge can only support 11 pounds before it collapses instantaneously.

Using the concept of centrifugal force (ie sand in a bucket) is there a speed that you can roll the sphere such that ...[text shortened]... as the sphere travels along the bridge?

Make any assumptions you need to that are realistic.

The centrifugal force is not important. It's net weight is still 12 pound. - 05 May '08 17:12

Are you sure?*Originally posted by FabianFnas***No, the bridge will not hold.**

The centrifugal force is not important. It's net weight is still 12 pound.

Let's assume that we really reach a situation where to c(centrifugal)-force spreads the sand evenly against sphere's inside. Now at every moment a part of all sand is being pushed upwards by the c-force. Now the net weight is a vectorial sum of all forces involved, including the force pushing the sand upwards. Mass remains the same, but some of the weight is canceled out. After all, weight is a force, isn't it? I'm not saying a whole pound can be canceled out, but at least by a little bit, the weight can be decreased.

P.S.

This is quite tricky to me but actually, if the c-force is dispersed evenly, it then cancels itself out and the weight remains.

Eh, gotta go learn some more dynamics. - 05 May '08 19:43

Yes, I'm sure.*Originally posted by kbaumen***Are you sure?**

Let's assume that we really reach a situation where to c(centrifugal)-force spreads the sand evenly against sphere's inside. Now at every moment a part of all sand is being pushed upwards by the c-force. Now the net weight is a vectorial sum of all forces involved, including the force pushing the sand upwards. Mass remains the same, but some ...[text shortened]... it then cancels itself out and the weight remains.

Eh, gotta go learn some more dynamics.

12 pounds is always 12 pounds. The net force down is 12 pounds. If the bridge will hold only 11 pounds it will crack. - 06 May '08 13:23

Yes, if it's even distributed then there's no change.*Originally posted by kbaumen*

This is quite tricky to me but actually, if the c-force is dispersed evenly, it then cancels itself out and the weight remains.

There are some potentially interesting cases if the distribution isn't even, though. Forget it's sand (more complicated) and just use a solid sphere with one half heavier than the other.

The average down force will be unchanged, but as the heavier side of the ball moves up and down (and therefore accelerates) the downward force will fluctuate. Roll it fast enough and the minimum force will be less than 11 pounds.

Now, this will only help you if you have a very short bridge (or very large ball!). But if the circumference of the ball is greater than twice the bridge length, and you time the roll correctly, you could probably arrange for the downward force to be below 11 pounds for the time it is in contact with the bridge.

Usual caveat: I think! - 06 May '08 13:58

But the sphere has sand inside: "We pour 2 pound of sand inside the hollow sphere but do not fill up the entire void inside the sphere." You cannot simplify the problem in order to find a solution that you want. Sand is not homogenous, if you can fill it, it moves inside the sphere.*Originally posted by mtthw***Forget it's sand (more complicated) and just use a solid sphere with one half heavier than the other.** - 06 May '08 14:19

I know. I was simplifying the situation for the sake of the argument. With sand the same principles apply, but it's a lot harder to calculate the exact effect. The one thing I'm pretty sure of is that you can't guarantee the force will remain constant in time.*Originally posted by FabianFnas***But the sphere has sand inside: "We pour 2 pound of sand inside the hollow sphere but do not fill up the entire void inside the sphere." You cannot simplify the problem in order to find a solution that you want. Sand is not homogenous, if you can fill it, it moves inside the sphere.** - 06 May '08 18:40 / 1 edit

But the risc is that you oversimplify the problem so you get a solution, but the wrong solution.*Originally posted by mtthw***I know. I was simplifying the situation for the sake of the argument. With sand the same principles apply, but it's a lot harder to calculate the exact effect. The one thing I'm pretty sure of is that you can't guarantee the force will remain constant in time.**

You cannot know at which side the sand will be so the problem turns into a probabilistic kind of solution like: "The chance that the bridge will hold is xx%". And from that no minimum velicioty can be calculable.

I say that it is highly likely that the bridge will collaps. - 06 May '08 22:04

No, it's entirely deterministic. But the solution will depend on the initial configuration and the history of the motion (e.g, how you accelerate the ball from rest). I've demonstrated a solution would almost certainly exist for some configurations - all you need is enough fluctuation in the down force - but calculating the exact parameters would be*Originally posted by FabianFnas***But the risc is that you oversimplify the problem so you get a solution, but the wrong solution.**

You cannot know at which side the sand will be so the problem turns into a probabilistic kind of solution like: "The chance that the bridge will hold is xx%". And from that no minimum velicioty can be calculable.

I say that it is highly likely that the bridge will collaps.*extremely*complicated. - 07 May '08 00:23 / 1 editI'd imagine that it would constantly exert 12 pounds. If we assume that centrifugal force is constant in every direction (on a plane perpendicular to the bridge and parallel to the sphere's bath), then the amount of decreased force caused by sand pushing upwards would be directly offset by the extra downward force from the other half of the sand. And even if all the sand is all at the top at a given point in time for some reason, then unless the sphere is unreasonably large (as noted by other posters), as the sand moved upwards it must have exerted an even greater downward force than its weight (e.g. in the range of 13-14 pounds total). Therefore even if there are times that it exerts less than 12 pounds of force downwards, it must have exerted at least that amount previously.
- 07 May '08 05:16

If you know every position and initial velocity of each grain (impossible), and... I stop there, I think you know where I'm getting at. Yes, by Newtonian mechanics, in theory, it is possible to calculate, but not in reality. So we have to rely on statistics. And then the anser is given above - the bridge will not hold.*Originally posted by mtthw***No, it's entirely deterministic. But the solution will depend on the initial configuration and the history of the motion (e.g, how you accelerate the ball from rest). I've demonstrated a solution would almost certainly exist for some configurations - all you need is enough fluctuation in the down force - but calculating the exact parameters would be***extremely*complicated. - 07 May '08 08:48

I know what you're getting at. But it is possible to calculate, at least close enough. Engineers solve problems like that every day, using computational fluid mechanics and similar techniques - I used to know people who worked on this sort of problem. Obviously I'm not going to try and do that now!*Originally posted by FabianFnas***If you know every position and initial velocity of each grain (impossible), and... I stop there, I think you know where I'm getting at. Yes, by Newtonian mechanics, in theory, it is possible to calculate, but not in reality. So we have to rely on statistics. And then the anser is given above - the bridge will not hold.** - 07 May '08 09:25

I refer you to my earlier post.*Originally posted by wolfgang59***I cannot believe that some of you are being suckered into this anti-grav invention!!**

The ball weighs the same when moving as it does at rest. The downward force is the same.

No anti-gravity, no change in weight. Just a fluctuating acceleration due to a rotating non-homogeneous body . The downward force only equals the weight if there is no upwards acceleration.