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Homework Help: Mainly centripetal forces, please check (esp #4)

  1. Jan 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Sasha's favorite ride at the fair is the Ferris wheel that has a radius of 7.0m.

    1. If the ride takes 20.0s to make one full revolution, what is the linear speed of the wheel?

    2. What centripetal force will the ride exert on Sasha's 50.0kg body?

    3. In order for Sasha to feel weightless at the top of the ride at what linear speed must the Ferris wheel turn?

    4. At this speed, how much will she appear to weigh at the bottom of the Ferris wheel?

    use 10 m/s^2 for gravity

    2. Relevant equations

    T = 1/f
    f = 1/T

    speed = (2)(pi)(r)/period

    centripetal acceleration = (linear speed)^2 / radius

    centripetal force = mass * centripetal acceleration ------- mv^2 / r

    angular momentum = mvr

    3. The attempt at a solution

    1. v= 2pir/t
    v = (2)(pi)(7)/(20) = 2.199 m/s

    2. centripetal force = mv^2 / r
    Fc = (50)(2.199)^2 / 7 = 34.544 N

    3. centripetal force i think must equal the same as Sasha's weight

    (50)(v^2) / 7 = (50)(10)
    50v^2 / 7 = 500
    50v^2 = 3500
    v^2 = 70
    v = 8.366 m/s

    4. at 8.366 m/s, she will weigh 418.33 N because
    (8.366)(50) = 418.33
  2. jcsd
  3. Jan 15, 2008 #2
    i didnt check the math but 1,2,3 look good.

    number 4 however is incorrect, what you've calculated is Sasha's linear momentum.

    Sasha's apparent weight will be equal to the normal force exerted on her at the bottom fo the ride.
  4. Jan 15, 2008 #3


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    Homework Helper

    For part 4. Consider the at the bottom of the ferris wheel and then consider the direction of the weight and the centripetal force.
  5. Jan 15, 2008 #4
    so does that mean i have to find her normal weight then times it by the centripetal force?
  6. Jan 15, 2008 #5


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    No. Her weight acts downwards.The centripetal force acts upwards. The resultant of the two forces is her apparent weight
  7. Jan 15, 2008 #6
    Her weight downward is 500 N and her centripetal force is 34.544N so 500 - 34.544 = 465.465 N. is that the answer?
  8. Jan 15, 2008 #7
    Fnet = mv^2/r, where mv^2/r is the centripital force.

    There are two forces acting on Sasha, gravity and a normal force in opposing directions, so taking the radial direction as positive.

    Fn - Fg = mv^2/r

    Sasha's apparent weight will equal the normal force, so

    Fn = mv^2/r + Fg
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