Question

A roller-coaster car (mass = 988 kg including passengers) is about to roll down a track. The diameter of the circular loop is 21.5 m and the car starts out from rest 36.2 m above the lowest point of the track. Ignore friction and air resistance.

X is the how high the car is= 36.2 m

Y is the diameter of the loop= 21.5 m.

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

This may help- the car reaches the top of the loop at the speed of 17m/s.

Answer 1

by energy conservation

suppose speed of the car at the top point =v

loss in PE = gain in KE at the top point

mg*( x -y ) = 1/2*m*v^2

v^2 = 2*g*(x-y)

v = sqrt(2*g*(x-y) ) =sqrt( 2*9.8*(36.2-21.5)) = 16.9741

so speed at the top point = 16.9741 = 17 m/s

the force exerted on the car by the track at the top of the loop

ΣF = N+W

r= 21.5/2 = 10.75 m

mv² /r = N+W

N = mv² /r - w

= 988*17^2/10.75 - 988*9.8 =

force on top = 16878.72 N