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A roller coaster's force and energy?

  1. Oct 25, 2011 #1
    http://www.google.com/imgres?q=roll...art=0&ndsp=15&ved=1t:429,r:14,s:0&tx=39&ty=51

    I had to design a rollercoaster that some what looks like this.

    And I have to figure out forces at the top and bottom of the loop.

    Can you tell me if i got this right??

    The force of the top will be weight and normal force pointing down and

    bottom will have weight down and normal force up.

    Also, what does it mean to have force is zero at the top of the loop???

    Can there be potential energy and kinetic energy at the loop?
     
  2. jcsd
  3. Nov 11, 2011 #2
    there is potential and kinetic at work -

    I don't know if this is your normal force, but you should include the centripetal acceleration.

    zero force at the top means centripetal acceleration (pointed up) equals gravitational acceleration (pointed down).
     
  4. Nov 11, 2011 #3

    sophiecentaur

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    Science Advisor
    Gold Member

    The speed of the car at any point can be found by calculating the lost Gravitational Potential Energy from the release point to the height in question. Then you can say that the Kinetic Energy gained is equal to the GPE lost - that will give you the speed at any height. We're ignoring frictional losses here.

    The car will follow the circular loop as long as its weight force is less than (or equal to) the centripetal force needed to keep it moving around a circle of that radius. If the weight force is greater, then the car will follow a tighter curve than the track - i.e. come away from the track and become a free-fall projectile.

    If you are having a problem visualising this then draw some arrows showing the weight force and the force with which the car is pressing against the track. Obviously, (?) the weight force must be less than or equal to the force on the track or the car will fall.
    The limiting case is when there is actually no force on the car when it's at the top of the curve - it's barely touching.
     
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