Question

4. A roller-coaster with a loop-the-loop is designed so that the cars are a very strong spring that is compressed by the cars and then the cars are rele from rest for the launch. The track is frictionless except for a 12m long rough section before the loop. The coefficients of static and kinetic friction between the cars and the track in the rough section are 0.8 and 0.12, respectively. The loop has a radius of curvature of 18m. What is minimum compression the spring can have so that the cars make it around the loop without losing contact with the track at the top of the loop? The spring constant of the spring is 9,500N/m and the cars have a mass of 180kg. friction 12m

Answer 1

here,

length of frictional track , s = 12 m

coefficient of static friction , us = 0.8

coefficient of kinetic friction , uk= 0.12

radius of curvature , R = 18 m

spring constant , K = 9500 N/m

mass , m = 180 kg

let the speed required by car at the top to just without loosing the contact be u

equating the forces at the top

N = 0 = m * u^2 /r - m * g

u = sqrt(18 * 9.81) = 13.3 m/s

let the compression in the spring be x

using Work energy theorm

Work done by friction = final potential energy in the spring - (initial kinetic energy + initial potential energy)

- uk * m * g * s = 0.5 * K * x^2 - ( 0.5 * m * u^2 + m * g * (2R))

- 0.12 * 180 * 9.81 * 12 = 0.5 * 9500 * x^2 - ( 0.5 * 180 * 13.3^2 + 180 * 9.81 * (2 * 18))

solving for x

x = 4.02 m

the compression in the spring is 4.02 m