What Is the Maximum Radius for a Roller Coaster Loop to Limit G-Forces to 2.5?

AI Thread Summary
To determine the maximum radius for a roller coaster loop that limits g-forces to 2.5, the initial height of 50 m is used to calculate the velocity at the lowest point, which is found to be 31 m/s. The centripetal acceleration equation is applied, incorporating the net acceleration of 2.5g. By rearranging the equation, the maximum radius can be derived. The calculations show that the radius must be carefully managed to ensure passenger safety and comfort. Understanding these physics principles is crucial for designing roller coasters that maintain acceptable g-force levels.
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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
 
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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, \frac{mv^2}{r} +g=2.5g
 
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