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Post by Hans Lemurson on Nov 9, 2005 23:10:00 GMT -5
Suppose that you are standing on a scale, holding some juggling balls. If you then begin to juggle them, does your average weight increase, decrease or stay the same?
Since juggling may be assumed to be a periodic oscilation, you only really need to think about the force averaged over the time of one cycle.
This may or may not be a trick question, but the answer sure boggled my mind when i figured it out did make me feel stupid when I realized my math-error. Still an interesting question though.
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Post by Evilduck on Nov 11, 2005 0:35:17 GMT -5
Are we taking into account the force that the balls exert on your hand once you catch them? (this would add a tiny fraction to your apparent weight)
If so your average apparent weight should increase (because it increases when you throw the ball AND when you catch it).
If not I think it should decrease because you are only holding the balls for part of the time.
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Post by Archagon on Nov 11, 2005 1:03:06 GMT -5
Do the balls count as part of "your average weight"?
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Post by Evilduck on Nov 11, 2005 1:15:53 GMT -5
That is a good point. I was assuming they did, because if they don't holding them would increase your weight even if you only held them for small periods of time.
However, sine I must question everything, is there a possibility that juggling objects would reduce your weight somehow? They certainly can't subtract from your mass, but my previous conjecture that they would add to your weight both at takeoff and landing seems wierd to me now.
Shouldn't they subtract form your weight at one instance and add at the other to keep balance?
My logic is that when a cannon fires a projectile upward with great velocity the opposite reacition pushes the cannon into the floor, but when the projectile lands in the cannon the force of its fall pushes the cannon into the floor again (thus making the cannon seem heavier both times)
I need an outside observer to check my logic for flaws...
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Post by Random on Nov 11, 2005 1:18:02 GMT -5
i agree, i believe that it would add weight at both points, i cannot see how it wouldn't
however its been some time since i've taken physics, so i could potentially wrong, but so could anyone
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Post by Archagon on Nov 11, 2005 1:29:13 GMT -5
Sounds ~right to me.
Takeoff: +weight (conservation of momentum), - weight of ball Landing: +weight (inelastic collision), +weight of ball
In other words, don't juggle in quicksand. ;D
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Post by Hans Lemurson on Nov 11, 2005 15:41:55 GMT -5
The '0' point for this situation will be defined as the weight exerted on the scale when you are holding the balls stationary. There is fundamentally no difference between catching and tossing the balls, as in both situations you are exerting a force on then to change their trajectory from 'down' to 'up'. The two factors you want to consider are how long the ball stays in the air, and how much force you exert on the ball while you are holding it (the catch-throw). When the balls are stationary, the air-time is zero, and the force exerted on them is 'mg' (mass * gravity). F avg*t = δ elta(m*V) (change in momentum). While the ball is in the air, it is experiencing the force of gravity, so its change in momentum is 'mg'*'air-time'. In order to launch it back into the air, you have to exert an amount of force 'F' over the amount of time you are holding the ball. The more force you exert, the greater weight the scale reads, but the less time you have to exert the force to get the ball flying. And while flying, you do not have to exert any force on the ball. And remember, always double-check your math if things don't make sense: This may or may not be a trick question, but the answer sure boggled my mind when i figured it out did make me feel stupid when I realized my math-error. Still an interesting question though.
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