For the rest of this problem consider just the motion of the center of mass of t

Question

For the rest of this problem consider just the motion of the center of mass of the rider. Consider a standard half-pipe with h = 13 ft with a 1 ft vert. Consider a 5.5 ft person whose center of mass is at his midheight. He starts from rest at location What is his speed v2 at location if he does not crouch (his center of mass is at the same height relative to his feet)? Based on your answer in part (c) above, how high (what is the location of his center of mass relative to location will he go up the vert on the other side? If he wishes to go up beyond the edge of the vert, he will need to have greater kinetic energy when he is at location To do this, the rider will pump as he goes back and forth on the half-pipe several times (he will crouch down as he approaches the flat to lower his center of mass) to further decrease his potential energy then spring back up to convert his internal muscular energy to kinetic energy. Suppose the rider in this case wishes to rise above the edge of the vert so that the change in his center of mass position is 15 ft relative to the flat section of the pipe (location what speed does he need to have when he is at location

Explanation / Answer

c)

.5 x v^2 = gh = 32 x 13

v = 28.84 ft/s

d) thus the height reached in this posiotion = kinetic energy = potential enrgy

heigh reched = intial height = 13 ft ,he will just reach the top if there is no friction and other losses

e)

KE = mgh = .5 x m x v^2

change in height = 15 ft

.5 x v^2 = 32 x 15

v = 30.983 ft/s

height that can be attained in this case assuming there are no losses will be 15 ft above the position 2

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