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I am trying to understand the physics behind this drunken ball:

There is a wight inside the ball as he mentions, but where, at what position ?

http://www.grand-illusions.com/acatalog/Drunken-Ball-661.html

I took one plastic ball, punctured, and put 25% volume of water. It was behaving some what similar to this but not as drunken as shown here.

So my quick question is what is the physics behind this drunken ball, and where is the weight placed in the ball. (I would like to make one drunken ball ).

Some of the comments from the youtube video:

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Albert Falck 1 month ago
its just a Little heavier ball inside of the wooden ball. That's it.
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Hide replies
Devin McLeod 1 month ago
No that's not it...
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Albert Falck 1 month ago
the ball is pretty near an edge of the ball, not in the middle

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hompalai 6 months ago
U can make a similar one with a plastic ball. Fill half of it with sirup, and place a middle sized marble/iron ball inside.
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    $\begingroup$ I suppose the mass distribution would have to be towards the edges of the ball. (Not uniformly, though.) The water you used inside the ball is dynamic, whereas an ordinary solid mass distribution is stationary, and that is why the ball you made isn't the same kind of "drunk" as the other one! $\endgroup$ – Hritik Narayan Apr 21 '15 at 12:36
  • $\begingroup$ @HritikNarayan What do you mean by edges ? ball is symmetrical $\endgroup$ – gpuguy Apr 21 '15 at 12:40
  • $\begingroup$ By edges I meant areas which are closer to the surface of the ball than to the center. (Sorry, I think edge shouldn't have been the word to use.) $\endgroup$ – Hritik Narayan Apr 21 '15 at 12:45
  • $\begingroup$ OK. Then I will try to recreate this and post here. Thanks for the pointers. $\endgroup$ – gpuguy Apr 21 '15 at 12:59
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At the beginning of the video it is shown that the ball seems to roll towards one stable orientation. This would imply that there are no free moving parts inside the ball, but that the center of mass does not coincide with the center of the ball. In your case this can be done by glowing a solid object to the inside of the surface of the sphere. An easy way of doing this would to freeze the water inside your ball, however this is only a temporary solution.

The physics involved can be described with Euler's equations. The torque, applied to the sphere, can be found by adding the torque due to gravity when the center of mass and the point of contact with ground do not align vertically and static friction when assuming that the is no slip between the ball and the ground. You could also add some roll friction or kinetic friction in order to simulate energy losses.

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  • $\begingroup$ Thanks for your suggestions. the freezing trick worked almost. I did this experiment with two balls. One with 50% water freezed, another with 25% freezed water. I found that the ball with 25% water appeared more drunken than 50%. May be due the CG being far away from the geometrical Center. $\endgroup$ – gpuguy Apr 27 '15 at 7:01

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