Originally posted by uzlessOkay, now I've read the question again. I agree that you're not asking about the fastest speed, but the minimum speed of the 10 pund ball.
I have a 10 pound ball. I have to get it across a bridge but the bridge can only support 5 pounds of weight.
My solution is to roll the ball across the bridge so that the horizontal momentum of the ball will decrease the vertical pull of gravity enough so that the ball will make it all the way across the bridge.
What is the minimum speed that i have to ...[text shortened]... sume zero friction and assume the ball doesn't slow down or speed up after you've released it.
I think we have not got all the neccesary facts obout the bridge. You haven't said anything about if the bridge is flat or not. I assume therefore tha bridge is flat. Also that the bridge is long enough to make this qeustion interesting. 1 cm long bridge is not interesting.
How much is 10 pounds of weight? Is is about 5 kilograms, or am I far way out? If we're talking about 5 kg, do we really need a bridge? Never mind, the bridge perhaps is made of spagetti, it doesn't really matters.
In the real world, if the builder says that the bridge can support 5 pound, he says that he garantuee this, not that it will collapse with even the slightest weight over 5 pounds. Under a short period of time it will support for perhaps 10 pounds, perhaps not. We need to know what the builder says about this. But since you don't mention any garantuees, we can assume that 5.0001 pounds will brake the bridge into pieces.
So, assuming that the bridge is flat, and will collaps if it is stressed more than 5 pounds, I don't think there is no lower nor upper limit in velocity. It will break anyway.
Edit: Now I have read the enire thread. Some has the same ideas that I have, others not. But I still think that vital information is missing. There is no way to calculate the minimal velocity if you don't assume things by yourselves. I still say the bridge will collaps.
I disagree.
I think that it is a fairly advanced physics/mathematics problem, but that all relevant details are given.
Assume 0 friction. Assume see level, on earth. Forget about the shape and makeup of the deck. Regardless of the bridge design specs, solve how to reduce the downward force of the ball by exactly half, because of forward momentum perpendicular to the deck.
Besides the stated problem itself, AThousandYoung has provided all the relevant information to do the research and calculate the answer. (I've just lacked the time to get back to it)
G = ma = mv^2/r
v =(rG/m)^(1/2)
http://en.wikipedia.org/wiki/Uniform_circular_motion
From this circular-motion page, it explains that what we really need to calculate to solve this problem is the speed at which the Centripetal force become 5lbs-or-grams, rather than the simple full 10lbs-or-grams (i.e. 0 speed), nor the somewhat simpler 0lbs-or-grams (i.e. a stable orbit).
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'bridge builder" is something you should all tackle
software with an elemental structural basis for the principles of general mechanics
so many bits - so far to span - so much to carry
there are numerous stages and it is quite interesting to see some the solutions that virtually defy standard structural analysis
sorry - got carried away there for a bit
knowledge of the bridge and its elasticity allows for a dynamic analysis
this analysis allows you to extend your calculation to sometimes ridiculous levels to determine an answer that is in itself an hypothetical situation sometimes devoid of a 'real' value
bridge design in itself has unusual allowances for irregular loadings
30 % here , 50 % there, impact loading in one region or direction etc
it is quite probable that a 5 lb load rating actually carries a 10 lb load in the real world, with safety factors in mind... but i detract from the question and i apologise
to limit the load to 5 lb when it is in fact 10 lb requires either centripital force against gravity requiring speed- else a flexible structure that has time dependant properties in response
timber structures have ratios of time dependant properties standardised to 75% in the australian standards for design for quick actions of 5 seconds or less- coupled with safety factors of applied loading 50%, and 20 % for dead loads = 1.75 * (1.5 + 1.2)/2 ( average when each is of a similar magnitude) = 2.36 so a timber bridge designed to carry a 5 lb load will carry a 10 lb load if it traverses in less than 5 seconds ( in australia... ) standards vary around the globe so this is not quite true everywhere
steel structures rely on those horrible elastic dynamic reactions of force against deflection and resonant properties of the frame and the speed of application and these codified equations are simply terrible to work with
i hope it's a timber bridge......
and i hope i haven't wasted anyones' time