Roller Coaster Terminal Height
Question:
A 200 Kg roller coaster starts on level track and enters a loop
with just enough velocity that it stays on the track, inverted,
at the top of a loop. Then on exiting the loop it enters an
inclined section of track. The radius of curvature of the loop at
its top is 25 m. The top of the loop is 35 m above the level
starting track. Determine the height above the level loop entry
track to which the roller coaster can rise on the exit incline.
Answer:
In order to just stay on the track at the top of the loop, the
centripital acceleration must be equal to the acceleration due to
gravity. Centripital acceleration is equal to the square of the
velocity at the top of the loop, Vt divided by the radius. So
Vt2/25=9.8 or Vt2=245, or Vt=15.7
m/s. That gives us a kinetic energy at the top of the loop
of 24,500 Joules. This kinetic energy will appear as potential
energy above this 35 meter altitude at the end of the run when
the coaster reaches its maximum height on the exit incline. The
height associated with a potential energy of 24,500 Joules is
h=24,400/(m*g) = 24,500/(200*9.8) = 12.5
m. Adding this to the 35 at the top of the loop we get a
terminal altitude of 47.5 meters.
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