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## Homework Statement

the roller coaster starts with the first hill being 50 m high. When the coaster gets to the lowest point, starting from the first hill, it has already entered the circular loop. The passengers should be subject to a maximum of 2.5 g's.

Figure out the maximum radius the circular loop can have.

## Homework Equations

v = squ.root of 2gh

v = squ.root of Rg

## The Attempt at a Solution

What is the velocity at the bottom of the first hill?

Solution:

ET(TOP) = ET(BOTTOM)

KE + PE = KE +PE

(1/2)mv2 + mgh = (1/2)mv2 + mgh

(1/2)v2 + gh = (1/2)mv2 + mgh (The masses cancel out because it is the same

coaster at the top and bottom.)

(1/2)v2 + gh = (1/2)v2 + gh (Substitute the numbers at each location)

( 1\2 ) (0) + (9.8) ( 50 ) h = ( 1\ 2 ) v^2 + 9.8 ( 0 ) (The height at the bottom is zero because it is the lowest point when comparing to the starting height.)

= 31 m/s