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Center of gravity question

  1. Apr 8, 2010 #1
    A skateboarder with his board can be modeled as a particle of mass 73.0 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 A). The half-pipe is a dry water channel, forming one half of a cylinder of radius 6.20 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 5.70 m. Immediately after passing point B, he stands up and raises his arms, lifting his center of mass from 0.500 m to 0.880 m above the concrete (point C). To account for the conversion of chemical into mechanical energy, model his legs as doing work by pushing him vertically up, with a constant force equal to the normal force nB, over a distance of 0.380 m. (You will be able to solve this problem with a more accurate model described in a later chapter.)
    d. What is the work done on the skateboarder's body in this process

    (mv^2)/r; W=F delta r

    I calculated the normal force to be 2146.26N which is mg + (mv^2)/r. However, knowing the work equation, what force do I use to calculate this work? Also, I have the angle as 0.. is that correct? My answer seems to be wrong.
    2146.26 x 0.38m
  2. jcsd
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