A skateboarder, initially at rest at the top edge of a vertical ramp half-pipe, travels down the pipe, reaching a speed of 6.8m/s at the bottom of the pipe. Friction is negligible. Use the law of...

A skateboarder, initially at rest at the top edge of a vertical ramp half-pipe, travels down the pipe, reaching a speed of 6.8m/s at the bottom of the pipe. Friction is negligible. Use the law of conservation of energy to find the radius of the half-pipe.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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A skateboarder, initially at rest at the top edge of a vertical ramp half-pipe, travels down the pipe, reaching a speed of 6.8 m/s at the bottom of the pipe. The force of friction is negligible. Let the radius of the half pipe ramp be R and the mass of the skateboarder is M. At the top edge of the ramp where the skateboarder the kinetic energy is 0 and the potential energy due to the gravitational force of attraction of the Earth is M*g*R.

At the bottom of the pipe the potential energy is 0 and has been completely converted to kinetic energy. As the speed of the skateboarder at the bottom is 6.8 m/s, the kinetic energy is (1/2)*M*(6.8)^2.

Using the law of conservation of energy gives:

M*g*R = (1/2)*M*(6.8)^2

=> 2*9.8*R = (6.8)^2

=> R = 46.24/19.6

=> R = 2.359 m

The radius of the half ramp is 2.359 m.

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