Naked Science Forum
Non Life Sciences => Technology => Topic started by: woodyapex on 27/03/2008 22:32:44
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[diagram=327_0] the square weight weighs 300kg the round one weighs 300kg .
the round weight is thrown of the ledge .
from top to bottom it's 6 m .
if there is no wind drag or little friction on the lip of the ledge how fast will the square be traveling when it gets to the end ?.
a hard one i know but i'm confident one of you can help.
[diagram=328_0]
same two weights but the time the round one is attached to a pulley and th rope is attached to the end of the ledge so in effect the round drops 6 m but the square will travel 12 m as the weight drops .
can i ask at what speed will it be traveling in this scenario??
please woody and thanks
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My worry would be the amount of friction in your system.
Would you not think a trebuchet could achieve the same end with less friction?
There may be reasons why you wish to use a linear catapult, but you will have to make careful assessment of the points of friction (which would be far greater than the would be for a trebuchet).
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George, I believe woodyapex (treetop?) wants to launch a flying machine of some sort.
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My worry would be the amount of friction in your system.
Would you not think a trebuchet could achieve the same end with less friction?
There may be reasons why you wish to use a linear catapult, but you will have to make careful assessment of the points of friction (which would be far greater than the would be for a trebuchet).
i just used two different shapes so i could name them round and square ! the square will be on wheels so very little friction and at the point where the rope turns at a right angle will be a roller so very little friction there too.
another if you can help that would be great.
thanks "treetop " lol
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F = mg = 300*9.8
accelerating 600kg (assuming rope is not stretchable and no friction etc)
So Acceleration a = 300*9.8/600 = 4.9m/s/s (or g/2)
After 6m
v = sqrt(2*a*s) where s=6
v = sqrt(58.84) = 7.67 m/s
Alternatively you can use energy conservation
2m*v*v/2 = m*g*s
v = sqrt(g*s) = sqrt(9.8*6) = 7.67 m/s as before
2nd question...
Just using energy conservation is easier here.
You know that the energy gained is (as before) m*g*s
And you know that the square weight will have twice the velocity (v) of the round one which will result in the total final kinetic energy being m*v*v/2 + m*(2v)*(2v)/2 = 5*m*v*v/2
so 5*m*v*v/2 = m*g*s
v = sqrt(2*g*s/5) = 4.85m/s (approx) with the square weight at 9.7m/s
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thanks a lot graham . just what i needed to know . unfortunately at the minute we are not in the flugtag but i'm going to keep trying .
thanks for your time
woody
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My worry would be the amount of friction in your system.
Would you not think a trebuchet could achieve the same end with less friction?
There may be reasons why you wish to use a linear catapult, but you will have to make careful assessment of the points of friction (which would be far greater than the would be for a trebuchet).
Looks like a homework assignment to me! That's the reason for the odd setup.
We should charge for help with those.