# physics

posted by on .

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 - ,

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 - ,

so Wap will equal in terms of m and g this.
m ( (sqrt ( r*g )/ r ) + g )

• physics - ,

v^2 = 4 g r
so
v^2/r = 4 g

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