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Jim01
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Homework Statement
Public skateboarding parks often include a well in the shape of a half cylinder (half-pipe). The skateboarder's path traces out a semicircular arc, with the midpoint of the arc at the lowest point. When starting from rest at one of the upper edges of the arc (at height h), the horizontal speed attained at the bottom can be shown to be the same as for vertical free-fall, v=[tex]\sqrt{2gh}[/tex]. At the bottom point, by what factor does the normal force from the skateboard on the skateboarder's feet exceed his weight?
My comments:
The force diagram of the skater at the bottom of the half pipe in relation to an x-y axis with horizontal and vertical directions: Normal force is pointing straight up, toward the center of the half-pipe and weight is pointing straight down, away from the center of the half pipe. In a separate diagram, centripetal acceleration is pointing straight up, toward the center of the half-pipe and centripetal force is pointing straight down, away from the center of the half pipe.
Homework Equations
Summation of Forces on the x-axis = 0
weight (W) on the x-axis = 0
Summation of Forces on the y-axis = m (v[tex]_{2}[/tex]/r)
Weight (W) on the y-axis = -mg
Normal (N) = mg
The Attempt at a Solution
N - W = m (v[tex]_{2}[/tex]/r);
N -mg = m (v[tex]_{2}[/tex]/r);
N = mg(1 + 2h/r)
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