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

Suppose the vertical loop has a radius of 8.92 m. What is the apparent weight (Wap) of a rider on the roller coaster at the bottom of the loop? (Assume that friction between roller coaster and rails can be neglected. Give your answer in terms of m and g.)

- physics -
**drwls**, Sunday, March 20, 2011 at 1:37pm
Didn'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 2:52pm
so Wap will equal in terms of m and g this.

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

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

so

v^2/r = 4 g

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

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