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Moment of inertia of solid slab

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Solid slab with moment of inertia 2 kg m2 is given the forces as shown.
    1-1.jpg

    Find the angular acceleration if θ = 37°, R1 = 10 cm, R2 = 25 cm

    2. Relevant equations
    τ = I . α

    3. The attempt at a solution
    This is what I've tried although I don't know it is correct or wrong; I tried to find all the torques by the 4 forces.
    τ1 = 10 x R2 (clockwise)
    τ2 = 10 x R1 (anti-clockwise)
    τ3 = 20 x R1 (anti-clockwise)

    I can't find the fourth torque because I don't know how to find the perpendicular distance from the inclined-10 N force to the center. Please help

    Thanks
     
  2. jcsd
  3. Jan 4, 2012 #2
    Hint: You can resolve it into 2 components: tangential and radial
     
  4. Jan 4, 2012 #3
    Sorry I don't get it. The arrow of the force is "toward" the slab, not "away" from it so if I resolve the force, I will get 10 cos θ and 10 sin θ but I don't see that one of them is tangential and one of them is radial.
     
  5. Jan 4, 2012 #4
    You are right, one of them wont actually be radial

    attachment.php?attachmentid=42453&stc=1&d=1325703328.jpg

    But since the blue distance is no given so i assume that its the imperfection of the figure causing the problem.
    If you have a way of finding the blue distance the its good or else IMO you can call it radial ...
     

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  6. Jan 4, 2012 #5
    OK I think I get it. Thanks for your help :)
     
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