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April 18, 2014

April 18, 2014

Posted by **HELP!** on Sunday, March 20, 2011 at 10:21am.

- physics -
**drwls**, Sunday, March 20, 2011 at 1:37pmDidn't you ask this yesterday? As BobPursley answered then, you need to know the velocity at the bottom of the loop. You still have not provided that information.

You can make a calculation of that velocity by using conservation of energy IF the velocity is zero at the top of the loop. They may expect you to make that assumption.

V(@ bottom)= sqrt(2gH) = 2 sqrt(gR)

Wap (@bottom)

= M (V^2/R + g)

= 5 M g

It is somewhat surprising that it is independent of R.

I have seen rides like that at carnival amusement parks, but never went on one.

- physics/ help -
**please recheck my answer again**, Sunday, March 20, 2011 at 2:52pmso Wap will equal in terms of m and g this.

m ( (sqrt ( r*g )/ r ) + g )

- physics -
**Damon**, Sunday, March 20, 2011 at 5:15pmv^2 = 4 g r

so

v^2/r = 4 g

so weight = m(4g+g) = 5 m g

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