The Sidewinder is a shuttle type steel rollercoaster. The rollercoaster cart starts at the bottom of the frictionless track and is accelerated from a stand still over a distance of 7 meters. Then the cart enters the loop, rises to the top with just enough velocity to keep from stalling at the top of the loop. Gaining velocity the cart finishes the loop, bottoms out, and then rises up the vertical segment of the track After stalling out at a maximum height the cart then return its path back to the starting point. This is a frictionless track with the radius of the loop measuring 7 meters. The mass of the cart filled with riders weighs 4000 kg. Determine the minimum speed the cart must have at the top of its are so that the cart continues moving in a circle. Calculate initial velocity (V_0) that must be obtained to have the velocity needed at the top of the loop. What is the acceleration of the cart that must be obtained in \"g\'s\"? What is the Kinetic Energy of the cart as it enters the loop? Top of the loop? What is the Kinetic Energy as it exits the loop? How high up the vertical climb does the cart climb?
a)
At the top of the arc, cart weight is balanced by centrifugal force.
mg = mv^2 /r
v = sqrt (rg)
= sqrt (7*9.81)
= 8.287 m/s
b)
KE at top = 1/2*mv^2
= 1/2*4000*8.287^2
= 137340 J
PE at top = mgh
= 4000*9.81*(2*7)
= 549360 J
Total energy at top of loop = mgh + 1/2*mv^2 = 137340 + 549360 = 686700 J
Initial KE = 1/2*m*Vo^2 = 1/2*4000*Vo^2 = 2000*Vo^2
Initial PE = 0
Total initial energy = 2000*Vo^2 + 0 = 2000*Vo^2
By energy conservation:
2000*Vo^2 = 686700
Vo = 18.53 m/s
c)
Using v^2 = u^2 + 2as we get
18.53^2 = 0 + 2*a*7
a = 24.525 m/s^2
= 24.525 / 9.81 g\'s
= 2.5 g\'s
d)
As found earlier, KE of the cart as it enters the loop = 1/2*m*Vo^2 = 686700 J
As found earlier, At top of loop, KE = 137340 J
e)
KE at exit of loop = KE at entry = 686700 J
f)
Again using, v^2 = u^2 - 2gh we get
0 = 18.53^2 - 2*9.81*h
h = 17.5 m
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