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Ferris Wheel Weight of the Rider

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A Ferris wheel has a radius of radius of 15.0 m. It rotates at a constant angular speed, and makes one full revolution in 25.0 s. Calculate the apparent weight of a rider (mass = 60 kg) when at the bottom of the wheel.

    r=15
    T=25s
    v=(2*pi*15)/25
    a=((2*pi*15)/25)^2/15

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    F=60*((2*pi*15)/25)^2/15=0.9465

    Am I doing it right or do I just multiply the mass by g?
     
  2. jcsd
  3. Oct 23, 2013 #2

    gneill

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    Staff: Mentor

    The rider will be feeling the effects of both gravity and the centripetal acceleration. That is to say, in your free body diagram (which I'm sure you've drawn, right?) the normal force provided by the seat of the Ferris wheel on the rider must counter the force due to gravity on the man's mass as well as provide the centripetal force to move that mass in circular motion.
     
  4. Oct 24, 2013 #3
    Alright so I have already solved for centripetal force. So I simply subtract that from the normal force?

    Which turns out to be this. (60*9.81)-0.9465=587.654N
     
  5. Oct 24, 2013 #4

    gneill

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    Staff: Mentor

    The rider should feel heavier at the bottom, not lighter. The normal force increases because it supports his weight due to gravity AND provides the centripetal force for circular motion.

    Also, recheck your calculations for the centripetal force. The final value does not look correct (it should be quite a bot larger).
     
  6. Oct 24, 2013 #5
    I recalculated centripetal force and got 56.79N. Adding that to the riders weight I get (60*9.81)+56.79=645.39 N.
     
  7. Oct 24, 2013 #6

    gneill

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    Staff: Mentor

    Looks good :smile:
     
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