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Physics merry-go-round problem

  1. Mar 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A small package is placed on the platform of a non-rotating merry-go-round, 2.72 m from the vertical rotation axis. At t=0, the merry go round begins rotating with a constant angular acceleration of α = 0.100 rad/s^2. The coefficient of static friction is 0.600. Determine the time t at which the package first begins to slide.

    2. Relevant equations
    ω = ωinital + αt

    3. The attempt at a solution
    My thinking of it was that the package would begin to slide after the max of its static friction is reached (μN), but was having trouble applying that and also expressing my answer in terms of time that has passed.
     
  2. jcsd
  3. Mar 7, 2016 #2

    BvU

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    Hello James, :welcome:

    So until it slides it makes a circular motion. What force is required to keep it in this circular trajectory ? (Magnitude in terms of ##\omega##?)
     
  4. Mar 7, 2016 #3
    Yes you are correct that the object will begin to slide once the maximum of the static friction is reached, but what force can you think of that will cause the object to start sliding?
     
  5. Mar 7, 2016 #4

    BvU

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    First things first, please.
     
  6. Mar 7, 2016 #5
    The centripetal acceleration will be what is keeping it in it's circular trajectory: a = v^2/r = ω^2 r, so in terms of omega: ω = sqrt(a/r)
     
  7. Mar 7, 2016 #6

    BvU

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    Good. What is providing the force that results in that centripetal acceleration ?
     
  8. Mar 7, 2016 #7
    Would it the spinning of the merry-go-round with the constant angular acceleration?
     
  9. Mar 7, 2016 #8

    BvU

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    Like: the more it spins the more centripetal force it provides ?
     
  10. Mar 7, 2016 #9
    I think it would be the faster it spins the more centripetal force it will provide
     
  11. Mar 7, 2016 #10

    BvU

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    Suppose you would be sitting there and it would be a bit slippery...
     
  12. Mar 7, 2016 #11
    After it starts constantly accelerating, eventually there will be a point where the velocity is high enough to overcome the friction and cause me to slide around on it
     
  13. Mar 7, 2016 #12

    BvU

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    Yes, in a way. It is the friction that keeps you in a circular trajectory. No friction and you start sliding outwards straightaway.

    So you have an expression for the force required to keep the package in a circular trajectory, and you have an expression for the maximum force the friction can provide...
     
    Last edited: Mar 7, 2016
  14. Mar 7, 2016 #13
    Ah, I see. So the centripetal acceleration is keeping the merry-go-round in motion, but it is the friction that is keeping the package in motion and not moving.
    And for the circular trajectory, its ω = sqrt(a/r) and the maximum friction in μ * N (normal force).
     
  15. Mar 7, 2016 #14

    BvU

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    Oops, no, there is some electric motor underneath that provides the torque for uniformly accelerated angular motion. The merry-go-round itself is a solid construction that stays together and thus provides its own centripetal force.
    The friction is keeping the package moving on a circular trajectory !

    It only appears not to be moving if you are on the merry go round: it does not move with respect to the merry-go-round.

    Can you manipulate your expression ##\omega = \sqrt {a/r} ## in such a way that you get the expression for the force required to keep the package in its circular trajectory ?
     
  16. Mar 7, 2016 #15
    The force would just be the acceleration of the merry-go-round? which would be a=v^2/r or ω^2 * r
     
  17. Mar 7, 2016 #16

    BvU

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    For a heavy package a greater force is required than for a light package. I don't see that in ##\omega^2 r## :rolleyes:
     
  18. Mar 7, 2016 #17
    Haha, I guess not. I mean, would it have something to do with other forces acting on it? Like gravity (mg) and the normal force from the ground of the merry-go-round (N)?
     
  19. Mar 7, 2016 #18

    BvU

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    No, what I mean is the ##\omega^2r## has the wrong dimension. It is an acceleration. According to good old Isaac, Force = mass times acceleration. So you need a factor ##m## in there !
     
  20. Mar 7, 2016 #19
    Ohhh, obviously. So something like ∑F = m * ω^2 * r
     
  21. Mar 7, 2016 #20

    BvU

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    Bingo. This is the force that is needed to keep the package in a circular trajectory. And ##\omega## is known as a function of time ##t##.

    The only force that can do that is the friction force with the floor, and you already had an expression for the maximum friction force ##F_{\rm fric, \ max} = \mu N## (to be worked out in known variables -- again something with the mass ##m## of the package :rolleyes:)

    As soon as required force > max friction the package can no longer be kept in a circular orbit !
     
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