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Rotational motion: playground spinning disk problem

  1. Sep 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A child is at a playground, and chooses to try the spinning disc (see figure). The radius of the disc is 2.00 m, and the coefficient of static friction between child-surface and disc-surface is μs = 0.350.
    In the following questions, you must provide algebraic equations as well as final numbers.

    a) If the child sits 1.50 m from the centre, how fast can the spinning disc turn, without her slipping off?

    b) The evil big brother now spins the disc with ω = 2.00 rad/s. At what distance from the centre of the disc should the kid sit, to avoid falling off?

    Having safely moved to the radius 0.500 m, the child starts complaining, becau- se she cannot get off the ride. The evil big brother decides to stop the rotating disc, by providing a constant tangential force.

    c) How quickly can he stop the disc, without her sliding in any direction (neither forwards, backwards, inwards or outwards)?


    2. Relevant equations (possibly)
    1. m⋅ν2/r (centrifugal force)

    2. ω = ν/r (angular velocity)

    3. a = -ν2/r (accelration)

    3. F = ma (Newton Law)


    3. The attempt at a solution

    Relevant information:

    r = 2.00m (radius of the disc)
    μs = 0.350 (static friction)
    Fs = μs⋅mg

    ra = 1.50m (distance from centre - in a)

    My attempt:

    2/r = μsmg

    μs = νmax2/rg

    vmax = √μsrg

    vmax = √0.350⋅1.50m⋅9.80m/s2 = 2.27 rad/s
     
    Last edited: Sep 26, 2016
  2. jcsd
  3. Sep 26, 2016 #2

    gneill

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    Hi ChrisBrandsborg,

    Your homework help request might get more and better response if the title reflected the nature of the problem and the are of physics involved. I've changed the title to: Rotational motion: playground spinning disk problem.

    You should be able to list a number of relevant equations including those related to rotational motion, centripetal acceleration, and friction. If you can't list them then perhaps you need to consult your text or course notes to review the topics. We really can't offer any help until you've shown that you've made an effort to research and attempt the problem.
     
  4. Sep 26, 2016 #3
    Hi! I did write some relevant equations, as well as an attempt to solve it, but I am not sure about everything! Could really use some help!
     
  5. Sep 26, 2016 #4

    gneill

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    In (a) the child sits at a radius of 1.50 m, so the centripetal force required will correspond to that radius. The speed of the disk should be specified in terms of rotational speed: radians per second (ω).

    You may find it easier to work in angular quantities rather than translating back and forth to linear ones. Centripetal force, for example, is given by mω2r, bypassing the need to convert to linear speed to use mv2/r.
     
  6. Sep 26, 2016 #5
    Yes, I forgot that it was rad/s. So the correct answer for a is: 2.27 rad/s ? :)
     
  7. Sep 26, 2016 #6

    gneill

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    Nope. You can't just change the units on a velocity to make it angular velocity. How do you convert a linear velocity to an angular velocity?

    Please don't keep changing your initial post. It makes the thread confusing to others when the followups don't reflect what came before.
     
  8. Sep 26, 2016 #7
    Oh, okay, so I use mω2r = μsmg, and solve for ω?
     
  9. Sep 26, 2016 #8

    gneill

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    That should do it :smile:
     
  10. Sep 26, 2016 #9
    Okay, so:

    ωmax = √(μsg/r) = 1.51 rad/s

    Correct?
     
  11. Sep 26, 2016 #10

    gneill

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    Yes. Looks good.
     
  12. Sep 26, 2016 #11
    I tried b as well:

    Now ω = 2.00, and we want to find r.

    μsmg = mω2r

    r = μsg/ω2 = 0.858m

    Correct? :) Thanks a lot for taking your time!
     
  13. Sep 26, 2016 #12

    haruspex

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    Yes.
     
  14. Sep 26, 2016 #13
    Then c:

    r = 0.500m
    Constant tangential force Ft
    ωstop = 0 rad/s
    t = unknown

    How do I find out how quickly he can stop the disc, without her sliding off?
     
  15. Sep 26, 2016 #14

    haruspex

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    Why would it be constant?
     
  16. Sep 26, 2016 #15
    "The evil big brother decides to stop the rotating disc, by providing a constant tangential force"
     
  17. Sep 26, 2016 #16

    haruspex

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    Ok, I missed that.
    What would be the angular deceleration?
     
  18. Sep 26, 2016 #17
    a = (ωf - ω0)/t ?
     
  19. Sep 27, 2016 #18

    haruspex

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    No, I mean given an applied tangential force Ft.
     
  20. Sep 27, 2016 #19
    I am not sure what Ft is.. Is it: Ft = mg⋅sinθ?
    So that at = gsin θ

    if we find a, can we then use:

    a = Δω/t, to find the time?

    or is it at = rΔω/t
     
    Last edited: Sep 27, 2016
  21. Sep 27, 2016 #20

    haruspex

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    It is the applied tangential force you are trying to find. I used Ft because that is what you called it in post #13.
    In terms of that unknown, what will the angular acceleration be?
     
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