Force of the track on a rollercoaster

  • Thread starter Rasine
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  • #1
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a roller coster car of mass 1500 kg starts at a distance of H=23m above the bottom of the loop 15 m in diameter. if friction is negligible, what is the downward force of the rails on the can when it is upside down at the top of the loop?

so what i was thinking is this:

f=ma and in this case a=v^2/r so i have .....

f=mv^2/r and i know what m and r are so i need to find v

ei=ef
mgh=mgh+.5mv^2
1500(9.8)23=1500(9.8)15+.5(1500)v^2

v^2=156.8

then i put this into f=mv^2/r and f=3.14 x10^4


is that right?
 

Answers and Replies

  • #2
Doc Al
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Note that f=mv^2/r gives the net force, but that you are asked to find the force of the rails on the car. (What other force acts on the car?)
 
  • #3
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the forces that act on the car are: normal force, weight, acceleration
 
  • #4
Doc Al
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Normal force and weight are the two forces acting on the car. Acceleration is not a force!
 

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