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Homework Help: Rotational Mechanics Problem again

  1. Dec 6, 2012 #1
    Hi friends the problem states like this:


    When the attached block is horizontal it is applying torque on the ring. Here I am using the formula T = Iα


    mg. r = (mr2(ring) + mr2(block))α

    so α = g/2r

    But it is not the correct answer. The correct answer is g/3r.

    Please help me friends. Thank you in advance.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 6, 2012 #2
    You need to take the torque due to friction force too.

    And check your moment of inertia for ring.
  4. Dec 6, 2012 #3


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    hey, man! What values have you used for the mass of the block and the mass of the ring?
  5. Dec 7, 2012 #4
    Sorry Bruce It happened by mistake.

    Taking mass of the ring as 3m the ans is g/4R.
  6. Dec 7, 2012 #5

    Hi Pranav.

    But here μ is also absent. Now about which axis I should take torque ? Because If I am taking it about the central axis μ is also there.
  7. Dec 7, 2012 #6
    The body rolls without slipping/sliding implies that there is force due to friction too.

    You have to take the torque about the CM of ring.
  8. Dec 7, 2012 #7


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    It is static friction, which needs to be enough to prevent the bottom point of the ring from sliding. No need to know μ.

  9. Dec 7, 2012 #8
    SO what should I do after that. What should I apply to solve it Not getting exactly. Please Help.
  10. Dec 7, 2012 #9
    Takinf torque about com,

    mg.r + μ. mg r = (mr2 + 3m r2). α

    now sticking with μ
  11. Dec 7, 2012 #10
    Now make an equation for the translational motion.
  12. Dec 7, 2012 #11
    Well I did some wrong in the previous equation,

    Right one -

    mgr + μ. 4mgr = (mr2 + 3mr2) α

    g + 4g. μ = 4r.α

    For translational motion,

    4μ mg = 3ma = 3 mrα

    Solving both equations,

    α = g/r

    Please tell me what i am doing wrong now.
    Thank you in advance.
  13. Dec 7, 2012 #12


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    hi thunderhadron! :smile:

    don't forget that the total centre of mass is moving downward, and the total moment of inertia is changing!

    you'll probably need to consider the ring and the mass separately :wink:
  14. Dec 7, 2012 #13
    I should have been a bit more clear in my previous post. You don't know the frictional force here, so instead of μ. 4mgr, write fr where f is the frictional force.

    For translational motion, f=ma=mrα (since its given that it rolls without slipping).

    Solve the two equations.
  15. Dec 8, 2012 #14


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    This time it is easier to use the instantaneous axis.

    Apply the parallel axis theorem to get the moment of inertia of the ring with respect to that axis, and add the moment of inertia of the small block.

  16. Dec 9, 2012 #15
    So as per you guys instructions I am doing it with full consciousness.

    Taking τ = Iα about the center of mass of the ring,

    mg.r + f.r = (3mr2 + mr2

    takinking F = ma for the com of the ring,

    f = 3m.a = 3mr.α

    solving both the equations I get the result,
    α= g/r.

    Which is not correct. That means still doing some wrong here. Please help.
  17. Dec 9, 2012 #16


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    What can be the direction of the force of friction? Think: when the ring rolls clockwise, the centre accelerates to the right. It is only friction that accelerates the ring and block to the right.

    The force of friction accelerates the whole system, ring and block. That is 4m.

    And I think the given answer (g/3r) is not correct.

    Last edited: Dec 9, 2012
  18. Dec 9, 2012 #17


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    I second echild's recommendation of using the instantaneous axis of rotation!
  19. Dec 9, 2012 #18


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    Anyway, I do not get the given result. Do you?

  20. Dec 9, 2012 #19


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    Me either. But I do get one of the other choices.
  21. Dec 9, 2012 #20


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    thirded! :smile:

    (and i too don't get the given answer)
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