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Rotation : When will the rod slide off?

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data
    one fourth length of a uniform rod of mass "m" and length "l" is placed on a rough horizontal surface and it is held stationary in horizontal position by means of a light thread. The thread is then burnt and the rod starts rotating about the edge. Find the angle between the rod and the horizontal when it is about to slide on the edge. The coefficient of friction between the rod and the surface is "μ"

    2. Relevant equations



    3. The attempt at a solution

    I have a few more questions in a similar format, and I cant find the right way to tackle them. Could someone please point me in the right direction?

    The answer as given in the book is θ= tan^-1 (4μ/13)
     
  2. jcsd
  3. Feb 7, 2012 #2

    tiny-tim

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    Hi AlchemistK! :smile:

    Start by using conservation of energy to find how dθ/dt depends on θ (before it slips). :wink:
     
  4. Feb 8, 2012 #3
    Ok, after conserving energy, I got:
    ω^2 = (24 g sinθ)/(7L)

    where θ is the angle made by the rod with the horizontal.

    On integrating, there comes a term of sinθ in the denominator and by applying the limits of θ from 0 to some angle θ, infinity comes up.

    The loss in PE came out to be m*g*L*sinθ/4
    The kinetic energy at any angle θ will be 1/2* I * ω^2
    where I is m*(L^2)/48.
     
    Last edited: Feb 8, 2012
  5. Feb 8, 2012 #4

    tiny-tim

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    ok, now do the equation for:

    Ftotal = mac.o.m

    (sum of the forces on the rod = mass times acceleration of the centre of mass)​
     
  6. Feb 8, 2012 #5
    In the direction of the rod,
    ma = mgsinθ - μmgcosθ
     
  7. Feb 8, 2012 #6

    tiny-tim

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    no

    you need to find the reaction force (call it R) at the table, from the acceleration

    then you check to see whether µ is capable of producing R

    (when it isn't, that's the answer)​
     
  8. Feb 8, 2012 #7
    Do you mean the frictional force on the table due to the rode? (μmgcosθ)

    Or the normal reaction due to the component of the weight of the rod? (mgcosθ)
     
  9. Feb 8, 2012 #8

    tiny-tim

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    are you sure that it's mgcosθ ?

    use the previous equation to find the total reaction force R (including its direction)
     
  10. Feb 8, 2012 #9
    Would that be the vector sum of mgcosθ and μmgcosθ? I cant see any more forces here.
     
  11. Feb 8, 2012 #10

    tiny-tim

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    No, it's the difference of ma and mg
     
  12. Feb 8, 2012 #11

    Aren't they numerically the same? ma - mg is equal and opposite to mgcosθ + μmgcosθ
     
  13. Apr 28, 2012 #12
    @tiny tim why did we find the relation b/w omega and theta
     
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