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Angular acceleration of a rod?

  1. Mar 31, 2007 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Itot= 1/3MrL^2 + 2/5 MsR^2 + Ms(L + R)^2 = 235.69784
    Cm= (Mr(L/2) + Ms(L+r))/(Mr+Ms) = 4.915

    angular accel = Torque/Itot
    Torque= 1/2Lmgcos(theta)

    3. The attempt at a solution
    I solved for Itot and Center of mass and got the correct answers but, I am having a difficult time getting the angular acceleration correct.

    Aa= .5Lmgcos(theta)/(Itot)
    = .5*6*4*9.8cos31/235.697
    = .42767 rad/s^2 which was incorrect
    Please help dont know what went wrong!
  2. jcsd
  3. Mar 31, 2007 #2


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    The 'L' in your angular acceleration formula should be the distance from the axis to the point where the force causing the torque is exerted. In this case the gravitational force can be treated as being exerted at the center of mass, not at the end of the rod.
  4. Apr 1, 2007 #3
    Do you have to account for the total mass of the system?
    .5*4.915*(4+4???)*9.8*cos31/235.697= .7 rad/s^2
  5. Apr 1, 2007 #4


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    4+4??? Don't you mean 4+1? Yes, total mass of system. And it looks like the 1/2 in your torque equation is coming from the assumption that the center of mass is the center of mass of the rod. Get rid of it and just put L to be distance from axis to center of mass.
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