- #1
user3
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A needle on a broken car speedometer is free to swing and bounces perfectly off the pins at either end, so if you give it a flick it is equally likely to come at rest at any angle between 0 and pi. If the needle has a length r, what's the probability density ρ(x) of the x-coordinate of the needle point - the projection of the needle on the horizontal line?
Answer: 1/(pi r sin(theta) ) = 1/(pi r √(1-(x/r)^2 ) )
I can't see why the answer is not simply 1/2r. The shadow of the tip of the needle is equally likely to be seen at any point from -r to r
Answer: 1/(pi r sin(theta) ) = 1/(pi r √(1-(x/r)^2 ) )
I can't see why the answer is not simply 1/2r. The shadow of the tip of the needle is equally likely to be seen at any point from -r to r