(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A skate boarder with height 1.5m and weight 60 kilograms

drops down a half-pipe with a radius of 4 meters using her

frictionless skateboard. At first, she carries her center of

gravity really low on the skateboard, say 1/4 her height.

At the bottom of the half-pipe, she pushes her center of

gravity up to 1/2 her height. She then continues up the other

side of the pipe. How much air does she catch (above the other side

of the half-pipe)?

2. Relevant equations

1/2mv2 + mgh = 1/2mv2 + mgh + Wapp

L = I Ω (omega)

Ω = v/r

I = m r

L = mr (v /r ) = mr (v /r )

r1v1 = r2v2

3. The attempt at a solution

what do i start with to derive to the following equation?!

d = (√(2g (r + h/4)) ( r-h/4)/(r-h/2))^2/ 2g

d = (r + h/4) (( r-h/4)/(r-h/2))^2

d = 4m + (1.5m/4)((4m-(1.5m/4))/(4m-(1.5m/2)))^2/(2 (9.81m/s^2))

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# Angular momentum and a skateboarder's halfpipe

Can you offer guidance or do you also need help?

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