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

[grigabuoy]

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

 

[chaoseverlasting]

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, $$\frac{mv^2}{r} +g=2.5g$$

 

[m2003]

Now, we can use the equation v = squ.root of Rg to find the maximum radius.

v = squ.root of Rg
31 = squ.root of R (9.8)
31^2 = R (9.8)
961 = 9.8R
R = 98.16 m

Therefore, the maximum radius of the circular loop should be 98.16 m in order for the passengers to experience a maximum of 2.5 g's. This is an important factor to consider in the design and safety of a roller coaster, as exceeding the maximum g-force can cause discomfort or even injury to passengers. By using the equations for kinetic and potential energy, we were able to calculate the velocity at the bottom of the first hill and use it to find the maximum radius for the circular loop.