A roller-coaster car shown in the figure below is released from rest from a heig

Question

A roller-coaster car shown in the figure below is released from rest from a height h and then moves freely with negligible friction. The roller-coaster track includes a circular loop of radius R in a vertical plane.

(a) First suppose the car barely makes it around the loop; at the top of the loop, the riders are upside down and feel weightless. Find the required height h of the release point above the bottom of the loop. (Use any variable or symbol stated above along with the following as necessary: g.)

h =

(b) Now assume the release point is at or above the minimum required height. Show that the normal force on the car at the bottom of the loop exceeds the normal force at the top of the loop by six times the weight of the car.

The normal force on each rider follows the same rule. Such a large normal force is dangerous and very uncomfortable for the riders. Roller coasters are therefore not built with circular loops in vertical planes. The photograph shows an actual design.

Explanation / Answer

a)
initial height h
height at top of circular loop = 2*R

use conservation of energy,
decrease in potential energy = increase in kineticc neergy
m*g*h' = 0.5*m*v^2
g*h' = 0.5*v^2
g*(h-2R) = 0.5*v^2
v^2= 2*g*(h-2R)
v = sqrt (2*g*(h-2R))

To feel weight less, Normal force = 0
so,
m*g must provide centrepetal force
m*g = m*v^2/R
g = v^2/R
g = 2*g*(h-2R) / R
R = 2*(h-2R)
R = 2*h - 4*R
2*h = 5R
h = 2.5*R
Answer: h = 2.5*R

b)
N at bottom, N1 = m*v1^2/R + m*g
N at top, N2 =m*v2^2/R - m*g

N1-N2 = m*v1^2/R - m*v2^2/R +2*m*g

use conservation of energy
Kinetic energy at bottom = potential energy at top + kinetic energy at top
0.5*m*v1^2 = m*g*h + 0.5*m*v2^2
0.5*v1^2 = g*h + 0.5*v2^2
v1^2 - v2^2 = 2*g*h

use this in
N1-N2 = m*v1^2/R - m*v2^2/R +2*m*g
N1-N2 = m (v1^2/R - v2^2/R) +2*m*g
N1-N2 = m (2*g*h/R) +2*m*g
use: h = 2.5*R
N1-N2 = m (2*g*2.5*R/R) +2*m*g
N1-N2 = m (5*g +2*m*g
= 7*m*g
=7*weight

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