Inch Mirror Ball
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A ball rolls down an inclined ramp onto a flat section of track.?
Near the end of the track, there is a hemishperical hill (x) inches high. The ball rolls up the hill, down the other side, and crosses a finish line with a total travel time of t(1).
The experiment is repeated with all variables identical, except that the hemishperical hill is changed to a mirror image hemispherical valley with depth (x). The ball rolls out onto the flat section of track, through the depression, and crosses a finish line with a total travel time of t(2).
Neglect friction. Assume that the diameter of the ball is very small compared to the diameter of the hemispherical hill and valley (they travel the same total distance). Assume that the ball never leaves the track.
Which ball finishes first?
The starting PE is
m*g*h
where m is the mass of the ball and h is the vertical height of release
in either case, both balls arrive at the edge of the hemisphere with the same KE, and therefore speed v, where
v=sqrt(2*g*h)
Intuitively I know that t2 One plausible argument will be to look at the ratio between speed of the roll with the hill at the top of the hill, versus the speed with the depression at the bottom of the depression. Note that they will both resume along with speed v=sqrt(2*g*h) on the flat part. For the hill for the depression A more complete argument will be to look at the difference between the travel time over the hill versus the travel time through the depression. Since the transit times over the rest of the track are identical, lets call t1 the time over the hill and t2 the time through the depression j
v1=sqrt(2*g*(h-x))
v2=sqt(2*g*(h+x))
v1/v2=sqrt((h-x)/(h+x))
clearly v2 is faster, so t2 is smaller.
the average speed over the hill is
(v+sqrt(2*g*(h-x))/2
and the average speed through the depression is
(v+sqrt(2*g*(h+x))/2
the displacement is 2*pi*x
so
t1=(4*pi*x)/(v+sqrt(2*g*(h-x))
t2=(4*pi*x)/(v+sqrt(2*g*(h+x))
clearly, t1>t2
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