You are designing the section of a roller coaster ride shown in the figure. Prev

Question

You are designing the section of a roller coaster ride shown in the figure. Previous sections of the ride give the train a speed of 13.1 m/s at the top of the incline, which is 39.1 m above the ground. As any good engineer would, you begin your design with safety in mind. Your local government?s safety regulations state that the riders? centripetal acceleration should be no more than n = 1.69 g at the top of the hump and no more than N = 5.61 g at the bottom of the loop. For this initial phase of your design, you decide to ignore the effects of friction and air resistance. What s the minimum radius you can use for the semi-circular hump? What is the minimum radius you can use for the vertical loop?

Explanation / Answer

let mass of the roller coaster is m kg.

consider the motion from the top (39.1 m height) to the ground level.

initial mechanical energy=kinetic energy+potential energy

=0.5*m*13.1^2+m*9.8*39.1

=469*m J

as there is no friction/air resistance in the whole ride, total energy at any point will be conserved.

now, on the top of the hump, let the speed is v.

then centripetal acceleration=v^2/Rhump

given that v^2/Rhump=1.69*9.8

==>v^2=16.562*Rhump......(1)

at the top of the hump, total mechanical energy

=0.5*m*v^2+m*g*height

=0.5*m*16.562*Rhump+m*9.8*Rhump

=18.081*m*Rhump

this energy is equal to 469*m J.

==>18.081*m*Rhump=469*m

==>Rhump=469/18.081=25.94 m

hence radius of the semi circular hump is 25.94 m

part b:

let at bottom of the loop , speed is v.

at this point , total energy=potential energy+kinetic energy

=m*g*0+0.5*m*v^2

=0.5*m*v^2

equating to total energy of the system,

0.5*m*v^2=469*m

==>v^2=938

==>v=30.627 m/s

hence centripetal acceleration at the bottom of the loop=v^2/R

given that maximum value is 5.61*9.8=54.978 m/s^2

then v^2/R=54.978

==>R=v^2/54.978

==>R=30.627^2/54.978=17.062 m

hence radius of the loop has to be minimum 17.062 m.

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