Roller Coaster Question

  • Thread starter grigabuoy
  • Start date
  • #1

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?
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

Answers and Replies

  • #2
Now that you have the velocity of the roller coaster, then you can figure out the centripital acceleration. According to the question, the net acceleration is 2.5g.

Therefore, [tex]\frac{mv^2}{r} +g=2.5g[/tex]