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Statics on an inclined plane

  1. May 17, 2017 #1
    1. The problem statement, all variables and given/known data
    On an incline with slope angle α there lies a cylinder with mass M, its axis being horizontal. A small block with mass m is placed inside it. The coefficient of friction between the block and the cylinder is μ; the incline is nonslippery. What is the maximum slope angle α for the cylinder to stay at rest? The block is much smaller than the radius of the cylinder.

    154ufdk.jpg


    2. Relevant equations


    3. The attempt at a solution
    Well, I'm quite sure that I must consider the torques of M and m (maybe the origin of my XY plane can be the contact point between the cylinder and the incline). In this case the torque ## T_{M}= R M g sin \alpha ## but how can I write the equation of ## T_{m} ## ?
    And is it correct to think that the two torque are opposite so until ## T_{m} > T_{M} ## the cylinder will stay at rest?
     
    Last edited: May 17, 2017
  2. jcsd
  3. May 17, 2017 #2

    TSny

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    If a line from the center of the cylinder to the small mass m makes an angle ##\theta## to the vertical, can you express the lever arm for the torque ##T_m## in terms of ##\theta## and ##\alpha##?
    Yes

    The cylinder stays at rest as long as the two torques have equal magnitudes. Also, you don't want the small mass to slip.
     
  4. May 17, 2017 #3

    CWatters

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    Perhaps remember that the slope of the curve on which mass m sits is orthogonal to the line between the centre point and m.
     
  5. May 18, 2017 #4
    Yes, but I don't know how to relate angle ## \theta ## with ## \alpha ##
    IMG_4395.JPG

    Because I have a isoscel triangle ( 2 side of length ## R ## ). Any hint?
     
  6. May 18, 2017 #5
    Is it possible that ## \theta = (90 - \alpha ) ##
     
  7. May 18, 2017 #6

    CWatters

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    θ is measured with respect to YR and YR is orthogonal to the plane X.
     
  8. May 18, 2017 #7

    TSny

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    To find the lever arm for ##mg##, it might help to identify the angle shown with the question mark below
    upload_2017-5-18_13-47-23.png
    I was thinking of ##\theta## as being the angle shown in red below
    upload_2017-5-18_13-55-39.png
     
  9. May 18, 2017 #8

    haruspex

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    They are related through the torque equation.
    How does θ relate to μ?
    (You will find it easier to work with TSny's definition of θ, as being the angle to the vertical; but either will work.)
     
  10. May 19, 2017 #9
    The angle ? is ##\alpha##


    Considering TSny's ## \theta ## angle ## \mu = tan \theta ## and m doesn't slide.
     
    Last edited: May 19, 2017
  11. May 19, 2017 #10

    haruspex

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    Right.
    Next, can you find the horizontal distance from the point of contact with the slope to the mass ? I see you already have the distance to the cylinder's mass centre line.
     
    Last edited: May 19, 2017
  12. May 19, 2017 #11
    But in order to find it I need the distance of the mass from my origin of axys XY (?)
     
  13. May 19, 2017 #12

    haruspex

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    what is the horizontal distance from the mass to the centre of the cylinder?
     
  14. May 19, 2017 #13
    Is ## R sin \theta_{1} ## with ## \theta_{1} ## the angle between Y axe and the mass
     
  15. May 19, 2017 #14

    haruspex

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    Right, and you know the horizontal distance from the point of contact to the cylinder's mass centre, so...
     
  16. May 19, 2017 #15
    No because I don't know ## \theta_{1} ##
     
  17. May 19, 2017 #16

    TSny

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    Does ##\theta_1## represent your choice of ##\theta## or my choice of ##\theta##? (I'm not sure if the Y axis here is the axis perpendicular to the slope or perpendicular to the ground.)

    Keep in mind that haruspex asked for the horizontal distance between ##m## and the center of the cylinder.
     
  18. May 19, 2017 #17
    My ##\theta_{1}## is different from your choice of ##\theta##

    Ohh well, the horizontal distance is ##R cos \alpha ##

    But I have to find the torque with respect to my origin of axys, haven't I?
     
  19. May 19, 2017 #18

    haruspex

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    Then I misunderstand how you are defining it. You wrote that it was the angle between the y axis (i.e. the verical) and "the mass". I took that to mean the radius that leads to the mass. That would make your angle the same as TSny's theta.
    No, that is the horizontal distance from the centre of the cylinder to the point of contact.
    Yes. And for that you need to find the horizontal distances from the point of contact to the two mass centres.
     
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