A skateboarder with his board can be modeled as a particle of mass 71.0 kg, loca
ID: 2176575 • Letter: A
Question
A skateboarder with his board can be modeled as a particle of mass 71.0 kg, located at his center of mass (which we will study in a later chapter). As shown in the figure below, the skateboarder starts from rest in a crouching position at one lip of a half-pipe (point circle a). The half-pipe is one half of a cylinder of radius 6.50 m with its axis horizontal. On his descent, the skateboarder moves without friction so that his center of mass moves through one quarter of a circle of radius 6.00 m.(a) Find his speed at the bottom of the half-pipe (point circle b).
(b) Immediately after passing point circle b, he stands up and raises his arms, lifting his center of mass from 0.500 m to 0.970 m above the concrete (point circle c). Next, the skateboarder glides upward with his center of mass moving in a quarter circle of radius 5.53 m. His body is horizontal when he passes point circle d, the far lip of the half-pipe. As he passes through point circle d, the speed of the skateboarder is 5.26 m/s. How much chemical potential energy in the body of the skateboarder was converted to mechanical energy in the skateboarder-Earth system when he stood up at point circle b?
(c) How high above point circle d does he rise?
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Explanation / Answer
PE_at top =KE_at bottom mgh =(1/2)mv^2 solve for v = sqrt(2gh) b) a = v^2/r c)Fc =mv^2/r N =Fc +W = m(v^2/r +g) Try to finish the rest