A) Find the location of the car just before launch. That is the distance d in th

Question

A) Find the location of the car just before launch. That is the distance d in the top view.

B) What is the maximum possible radius of the loop if the car must stay on the track (include FBD and work-energy diagram)

A new roller coaster uses a sling shot design to launch a car from rest. The car travels without friction through the loop-de-loop. After point D, the wheels lock and the car slides on the rails with a coefficient of friction of mu k = 0.4 before coming to a rest. Given that the speed of the car at point B is 80 m/s, answer the following questions: A) Find the location of the car just before launch. That is the distance d in the top view. B) What is the maximum possible radius of the loop if the car must stay on the track (include FBD and work-energy diagram) C) The coefficient of friction after point D is mu k = 0.4, how far will the car travel past point D?

Explanation / Answer

as the surface is smooth, speed at launch=speed at B

then energy in spring=kinetic energy of the roller coaster

0.5*40*80^2=0.5*2000*x^2

where x is the elongation

x=11.314 m

spring unstretched length=l0=5 m

sqrt(l0^2+d^2)=(11.314+5)

d=15.53 m

b)for maximum radius,at the top i.e. at C, the car should stick to the track

i.e. the gravitation force should just balance the centripetal force

hence m*g=m*v^2/r

=>v^2=g*r...(1)

now using energy balance between bottom point of the loop and top point of the loop:

0.5*40*80^2=0.5*40*v^2+40*9.8*2*r

putting v^2=g*r

we get r=593.69 m

c) at point D, speed=80 m/s
deceleration=0.4*9.8=3.92 m/s^2

so distance travelled=80^2/(2*3.92)=816.33 m

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