A roller coaster car (mass = 1,175 kg including passengers) is about to roll dow

Question

A roller coaster car (mass = 1,175 kg including passengers) is about to roll down a track. The diameter of the circular loop is 10 m and the car starts out from rest 63 m above the lowest point of the track. Ignore friction and air resistance.

(a) At what speed does the car reach the top of the loop?

(b) What is the force exerted on the car by the track at the top of the loop?

(c) From what minimum height above the bottom of the loop can the car be released so that it does not lose contact with the track at the top of the loop?

Explanation / Answer

a)
Beetween top of track and between top of loop,
decrease in height= 63-10 = 53 m

use conservation of energy = decrease in potential energy = increase in kinetic energy
m*g*h = 0.5*m*v^2
g*h = 0.5*v^2
9.8*53 = 0.5*v^2
v=32.23 m/s
Answer: 32.23 m/s

b)
Let that force be Normal force , N
then
N +mg = m*v^2/R
N + 1175*9.8 = 1175*(32.23)^2 / 5
N = 232597 N
Answer: 232597 N

c)
if N= 0
mg = m*v^2/r
v^2 = r*g

Let height be h
then height difference= h-10

m*g*(h-10) =0.5 m*v^2
g*(h-10) =0.5 v^2
g*(h-10) =0.5 *(r*g)
h-10 = 0.5*r
= 0.5*5
= 2.5 m
h = 12.5 m
Answer: 12.5 m

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