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Dynamics - Pulley and Rod

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data
    There is pulley which has a mass hanging from one side, and a rotating rod attached horizontally to the other (both by cables). The pulley has mass; Mp = 9.5kg and radius; Rp =0.2m. The block(A) has mass; Ma=10.9kg. The rod is L = 0.8m long and has mass=Ml7.9kg. The rod is also rotating at w= 2.7 rad/s in the anti clockwise direction. The value for gravity used is g= 9.8m/s/s downwards. Tensions are denoted by Ta and Tb

    2. Relevant equations
    Newtons 2nd Law(F=ma), Eulers Equation (M=I[itex]\alpha[/itex]).


    3. The attempt at a solution
    I have defined my coordinate system as anticlockwise is positive and vertically down is positive.
    Ipulley = 0.5MR2
    Irod at centre of grav =(1/12)ML2
    I have developed some simultaneous equations for the system:
    Ta - Mag = Maaa From Newton 2nd Law of Block(A)
    aa = Rp[itex]\alpha[/itex]p Acceleration relationship
    -TaRp+TbRp=Ip[itex]\alpha[/itex]p Eulers equation for pulley
    Tb - Mrg = Mrar Newtons second law on rod.

    I also have the following(which i think is where the problem is at):
    -TbL - Ir[itex]\alpha[/itex]r = -Lw2 Eulers equation on the rod at the centre of gravity.
    ar = -L[itex]\alpha[/itex]r

    I solved this system, and got an incorrect answer. As above i think the problem lies in the last two equations.
     
    Last edited: Nov 5, 2013
  2. jcsd
  3. Nov 4, 2013 #2

    Simon Bridge

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    Welcome to PF;

    OK - rod rotating anticlockwise means the pulley is rotating clockwise and the mass is moving up.
    The most likely location for an error is a misplaced minus sign as you link the different fbds.

    The signs in your equations don't seem to match your stated sign convention.

    i.e. ##T_a-M_ag=M_aa_a## gives a negative acceleration if the weight is greater than the tension.

    But you've set them up how I would except for the last one, I'd have picked anticlockwise for positive for the motion of the rod.... done that way the acceleration (magnitude and sign) of the block A is also the tangential acceleration of the rim of the pulley and the acceleration of point B on the rod.

    However - the problem statement lacks a goal.
    What are you supposed to be finding out?
     
  4. Nov 4, 2013 #3
    My apologies for not including the problems goal.
    The problem asks to find the acceleration of pint B in the diagram provided.
     
  5. Nov 5, 2013 #4

    haruspex

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    I too am confused about the signs.
    makes up +ve
    makes down and anticlockwise the same sign
    makes clockwise +ve
    makes up +ve
    Makes anticlockwise +ve and ... a couple of problems here.
    The rod is rotating about an endpoint so you need to use the parallel axis theorem.
    The Lw2 term is a centripetal acceleration, which at this point is horizontal. It is supplied by the axle and has nothing to do with the tension.
     
  6. Nov 5, 2013 #5
    Thanks guys!
    By working through the sign troubles and getting rid of the Lw^2 term, the solution was achieved. Your help is appreciated greatly, and i guess ill need to work on sign conventions before my exam.
     
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