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Re: [Phys-l] time to bottom of ramp



At 05:33 PM 10/27/2006, Krishna ., you wrote:
.... For
a frictionless quarter circle of radius h = 100 feet, I find the travel time
to be 3.27757 seconds ...
the quarter circle is oriented so that the initial part of the arc is
vertical (aligned with the uniform gravitational field) and that the end of
the arc is horizontal.

For an inclined plane of slope 1 with a vertical drop and horizontal
traverse each of 100 feet (ie an inclined plane between the initial point
and the final point of the quarter circle arc above), I find the travel time
to be 5/Sqrt(2) = 5.5355 seconds....
...
regards
-Krishna

Krishna Chowdary
Department of Physics & Astronomy
Bucknell University


It is interesting to explore limits...
if one starts with that vertical drop of 100 ft, and smoothly rotates
to a lossless continuation on a 100 ft horizontal plane
the time is 3root (200g) / 2g = 3.75 sec.
But a straight ramp from coordinates (0,100ft) to (12.85,61.18 ft)
followed smoothly by another straight one from coordinates
( 12.85, 61.18ft) to (100,0 ft)
takes 3.28 seconds....


Brian Whatcott Altus OK Eureka!