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Finding Apparent Weight with Uniform Circular Motion

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data

    A ferris wheel with a radius of 14.0 m rotates at a constant rate, completing three revolutions in 50.0 s. What is the apparent weight of a 60.0 kg passenger when she is at the top of the wheel?
    Given choices: 532 N, 452 N, 562 N, 625 N, 469 N

    2. Relevant equations

    T=2∏/ω, a=(ω^2)r, F=ma

    3. The attempt at a solution

    3 rev/ 50 s = 0.37 (rad/s)
    a=1.92 (m/s^2)
    F=(a+g)m=(1.92+9.81 (m/s^2))(60.0 kg)=703.8 N
     
  2. jcsd
  3. Apr 7, 2013 #2
    Your attempt is a bit incomplete. You are forgetting to include the force due to centripetal accelaration which is
    2R
    Find it and subtract it from the value u got in ur attempt
    The force equation on the man is
    m(g+a)-mω2r
     
  4. Mar 23, 2016 #3
    Firstly we should be clear about definition of weight ,as per halliday "The weight W of a body is the magnitude of the net force required to prevent the body from falling freely, as measured by someone on the ground"..Now what you have found in your attempt is weight at the lowest point on the wheel.At the topmost point it should be M(g-ω^2r)..think it like a body in an elevator moving downwards at acceleration of magnitude ω^2r. Mathematically you can remember weight as M(⃗g-⃗a). so at lowermost position ⃗g-⃗a will come out to be 9.8+ω^2r and at top 9.8-ω^2r. For this particular question answer is 469 N.
     
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