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Classical Mechanics-Moments of Inertia and Torques

  1. Dec 3, 2013 #1
    1. The problem statement, all variables and given/known data
    a)Two people are holding the ends of a plank of length l and mass M. Show that, if one suddenly lets go, the initial acceleration of the free end (aD) is 3g/2. (7 marks).



    Moment of inertia, I, of the plank about its centre of mass is given by I=1/12(Ml2)

    b)Show also that the load supported by the other person falls from Mg/2 to Mg/4. (3 Marks).
    2. Relevant equations
    I*angular acceleration=torque=rxF

    parallel axis theorem for the moment of inertia
    3. The attempt at a solution
    I have completed the first 7 marks with no difficulty but am really struggling with how to set part b) up and what assumptions I need to make. I've spent hours and hours on this and have ended up going round in circles.

    I think I may need to calculate the linear acceleration of the centre of the plank and from this deduce the new load on the other hand. However I have not made much progress with this part of the question.
     
  2. jcsd
  3. Dec 3, 2013 #2

    Doc Al

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    Staff: Mentor

    That's exactly what you have to do. What forces act on the body? Apply Newton's 2nd law.
     
  4. Dec 3, 2013 #3
    So I need to calculate the linear acceleration of the centre of the plank then apply newtons 2nd law to this? This would give the net force on the plank? After this I could deduce that the the net force=mg-(the load on the other hand? Would this work?
     
  5. Dec 3, 2013 #4

    Doc Al

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    Staff: Mentor

    You got it.
     
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