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Rotational Motion Problem - 2

  1. Dec 16, 2013 #1
    Hi friends,
    Please help me in solving this problem, I'll appreciate the help.

    The problem is as:

    https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1471110_1461726330721139_1078073816_n.jpg

    Attempt:

    https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn2/1486585_1461727017387737_725847173_n.jpg

    Thank you all in advance.
     
  2. jcsd
  3. Dec 16, 2013 #2

    haruspex

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    I don't think there's any shortcut here. Write out the usual force/torque/acceleration equations, assuming rolling contact and taking the point of application of F to be some height x above the centre of mass. It doesn't say whether the cylinder is uniform, so maybe put I as the Moment of Inertia for now.
     
  4. Dec 17, 2013 #3
    Taking torque about the centre point and let the cylinder to be uniform,

    F. (x) + f(R) = (I).α
    => F(x) + f(R) = 1/2 MR2
    =>Fx + fR = MR2 α/2 - - - - - - -(1)

    Force equation,

    F - f = ma (a be the acceleration of c.o.m.) and for no slipping, a = Rα

    So, F - f = mRα - - - - - - -- - (2)
     
  5. Dec 17, 2013 #4

    haruspex

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    So eliminate α to find f as a function of x.
     
  6. Dec 17, 2013 #5
    f = F - [2F(R+x)]/ 3MR2
     
  7. Dec 17, 2013 #6

    haruspex

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    No, that's dimensionally incorrect. Try it again.
     
  8. Dec 17, 2013 #7
    Yes, Sorry for this.

    The correct one is,

    f = F - [F(x + R)]/ 2R ; 0≤x≤R

    Here, x can not exceed R.

    For x = 0, f = F/2 i.e. translational motion is happening.
    For x = R, f = 0 i.e. pure rolling

    But for any value of x < R, f will have some positive value. i.e. it will act backwards.
     
  9. Dec 17, 2013 #8

    haruspex

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    Still not right. You should get a term like 3Rf.
     
  10. Dec 17, 2013 #9

    tiny-tim

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    hi coldblood! :smile:
    yes there is

    take torque about the bottom of the cylinder (the point of contact and centre of rotation)

    [you can do τ = Iα about a moving point provided

    i] it is the centre of mass

    or

    ii] it is the centre of rotation and the centre of rotation is moving in a straight line parallel to the centre of mass :wink:]​
     
  11. Dec 17, 2013 #10

    haruspex

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    I merely meant that you can't reason it out without getting into equations (as had been attempted in the OP). Yes, some routes to the equations are distinctly faster than others.
     
  12. Dec 17, 2013 #11
    are the equations which I wrote in post 3 right?
     
  13. Dec 18, 2013 #12
    Yes I got the correct equation,

    3Rf = F (R - 2x)
     
  14. Dec 18, 2013 #13

    haruspex

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    Good. Now, what is the range of possible values of x?
     
  15. Dec 18, 2013 #14
    3Rf = F (R - 2x)

    x lies between 0 to R
    when x = 0, f = F/3
    when x = R, f = - f/3
    when x = R/2, f = 0
     
  16. Dec 18, 2013 #15

    haruspex

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    Right. Can you select an answer now?
     
  17. Dec 19, 2013 #16
    Well the confusion which is arising is that I got the function

    3Rf = F (R - 2x)

    x lies between 0 to R
    when x = 0, f = F/3
    when x = R, f = - f/3
    when x = R/2, f = 0

    using the condition of pure rolling, I used a = Rα for the bottom point.
     
  18. Dec 19, 2013 #17

    haruspex

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    Right, and that does match one of A, B, C, D. Which one?
     
  19. Dec 19, 2013 #18
    Last one when x = R/2, f = 0, Because during pure rolling contact point does not slip. Hence friction is zero.
     
  20. Dec 19, 2013 #19

    haruspex

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    No, x can be anything from 0 to 2R. You need an answer that's valid for any x. If x is not R/2 there will be a frictional force.

    You said friction = 0 was the last choice. It's the third choice of four.
     
  21. Dec 19, 2013 #20
    I got it. That means Any of the choice could be there. Hence Can't be interpreted?
     
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