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Rotation: Speed of an object as it slips off a rotating disk

  1. Mar 26, 2014 #1
    1. The problem statement, all variables and given/known data

    A [itex]75g[/itex] mass sits [itex]75cm[/itex] from the center of a rotating platform undergoing a uniform angular acceleration of [itex] 0.125rad/s [/itex]. The coefficient of static friction between the mass and the platform is [itex]0.250[/itex]. What is the speed of the mass when is slides off?

    A. 0.889 m/s
    B. 1.26 m/s
    C. 1.44 m/s
    D. 1.58 m/s
    E. It will never slide off

    I'm also given an image of a disk with a mass on it, and the mass is located at the edge of the disk a distance r from the center.

    My variables as far as I can tell:
    α=0.125rad/s2
    m=75g
    r=.75m
    g=9.8m/s2

    2. Relevant equations

    ∑Fx=fs=mar
    ∑Fy=N-mg=0
    fssN
    ar2r
    at=αr

    I'm not sure if there are other equations I need in order to solve this problem

    3. The attempt at a solution

    I set up a free body diagram using the sum of the forces in the x and y directions:
    ∑Fx=fs=mar
    ∑Fy=N-mg=0

    I solved for fssmg

    I replaced a w/ ω2r to get ∑Fxsmg=mω2r

    Then I solved for ω to get [itex]ω=\sqrt{μg/r}=\sqrt{\frac{(.25)(9.8)}{.75}}[/itex]. This gave me an answer of [itex]1.807rad/s[/itex] which isn't close to any of the given answers.

    I'm also unsure because I didn't actually use the value for the angular acceleration. I thought that by solving the problem how I did I would be able to find the maximum speed the object would be able to rotate on the disk without slipping, and any values greater than that would cause the object to slide and fall off. I'm struggling with this problem and I don't know if I'm even starting correctly.
     
  2. jcsd
  3. Mar 27, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Hi sksmith6471. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]


    EDIT. Your answer has units of radians/sec. The answer options to choose from have units of m/s.

    The answer I get is not listed; it is partway between that of B and C.
     
    Last edited by a moderator: May 6, 2017
  4. Mar 27, 2014 #3
    Wow, thank you. I didn't even notice my units were different than the given answers. I converted my units to m/s using ##v=\omega r##. This gave me 1.356m/s which is between answers B and C. This should be the speed of the object just before it starts to slip.

    Did I approach this problem correctly? If the speed was even slightly greater than this value it should start to slip right? These questions are created by our instructor so there may be an error with the answers.
     
  5. Mar 27, 2014 #4

    NascentOxygen

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    Staff: Mentor

    Your approach looks right, and your answer matches mine.
     
  6. Mar 27, 2014 #5
    You're both missing the fact that the centripetal acceleration is just one of the components of the acceleration. There is also a tangential acceleration to be calculated. The total acceleration can then be found using Pythagorean theorem and related to the friction force using Newton's 2nd law.
     
  7. Mar 27, 2014 #6

    NascentOxygen

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    Staff: Mentor

    You're right! I originally calculated tangential accelerating force, saw that it was well below tangential friction so dismissed it. I think I can see which option to choose, now.
     
  8. Mar 27, 2014 #7
    I'm a bit confused. Are you saying that ∑Fx=fs=mar should actually be ∑Fx=fs=mα? then I take the squareroot ar2+at2 plug it in to my force equation and then solve for ω?
     
    Last edited: Mar 27, 2014
  9. Mar 27, 2014 #8
    Yes, that's right.
     
  10. Mar 27, 2014 #9
    Ok, I set up my equation: ∑Fx=fs=mα→ μmg=m\sqrt{ar2+at2} (I'm not sure why itex isn't working for me, so i omitted the commands). After some simplification I get ω2=\sqrt{(μg)2-(αr)2}/r.

    I plugged in my values at this point and found that:
    ω2≈3.264rad/s → ω≈1.806rad/s.

    Then i plugged in :
    v=ωr → v=(1.806)(.75)≈1.355m/s.

    Edit: I'm also confused as to why (assuming my math is correct) the answer that I just got is approximately the same as in my second post.
     
    Last edited: Mar 27, 2014
  11. Mar 27, 2014 #10
    Now your calculation seems correct. And you're right, the tangential acceleration had little effect because the angular acceleration is so small. Are you sure you provided the correct data for the problem?
     
  12. Mar 27, 2014 #11
    I spoke with my teacher and he said there was a typo in the answer selection. The correct answer is in fact 1.36 m/s and answer B should be 1.36 m/s. Thank you dauto and NascentOxygen.

    As a side note: he also mentioned that the acceleration he gave was the tangential acceleration. Luckily the value doesn't make too much of a difference so the answer wasn't affected by much.
     
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