Simple yet mind boggling!!!?

Question

Suppose there is a right isosceles triangular wedge with the hypotenuse replaced by 1/4th the circumference of a circle of radius R.
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Now there is a block of mass M that we release from its top. How do u find the time in which it comes down i.e a height of R(wedge is fixed).

hey what ru so confused with. u can consider the structure to be a quadrant of a square with side 2R. and a circle of radius R inscribed in it with the region containing the circle removed. im talking abt a cross section. i cant do better than that now.

Answer 1

If the curve is concaved, the Alexander -- as usual -- is indeed correct. However, the formula looks like the variables are upside down. (This is probably from forgetting a parenthetical). It should read:

T = 1.18 * (π/2) * √(R/g)

Then again, there is nothing keeping the mass from falling off the back side of the triangle. :-)

Note: if you inserted a convexed 1/4 circle to make it look like Half Dome (See http://en.wikipedia.org/wiki/Image:Yosemite_22_bg_090404.jpg ), the mass would be in unstable equilibrium at the top and you could not determine when it would start sliding.

Answer 2

I'm confused by the statement of the problem. Are we just talking about a particle sliding down the arc of a circle of radius R? That would seem infinite if you start it at the very top (it'll just sit there).

So I think you need to specify the width of the base of this wedge in terms of the radius R. I think you are saying the height of the wedge is R, right?

Added--

I thought I had it for a bit there, but now I'm confused again. I see no way for the height of the fall to be the same as the radius of the circle with 1/4 of the circumference replacing the hypotenuse without the particle starting on the flat top of the circle. Am I missing something? Can you draw a better picture?
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Aha -- you're talking about a circle that's concave outward. I thought you were describing something that curved the other way. Now I think I'm straight.

Answer 3

This problem is easily reduced to elliptic integral, and the final answer cannot be expressed in elementary functions.

This time is equal to quater of period of a pendulum of length R oscillating with amplitude π/2.
This time is approximately equal to T = π/2 √(g/R) * 1.18.

1.18 is the value of elliptic integral
1 + (1/2)² 1/2 + (3/8)² 1/2² + (15/48)² 1/2³ + ...