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A cyclist rolls down a ''devil's loop'' track from a height H. Find the pressure of the cyclist on the track as afunction of the angle the radius vector makes with the vertical. Do the calculations also for the case when the cyclist rolls down from the minimum height. |
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Answer» `N - mgcos alpha - mv^2//R` To find speed apply the law of conservation of energy . `mgH = mgR(1 cos alpha) + mv^2//2` Hence `N=mg(3cosalpha-2+(2H)/R)` In the highest point of the loop `alpha = pi` , so `N_("top") =mg (-5+(2H)/R)` The MINIMUM height is FOUND form the condition `N_("top")=0` , there FORE `H_(min) = 5/2 R`. In this case `N=3mg(1+cosalpha)=6mgcos^2""(alpha)/2` 
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