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Rotational Motion of a block of mass

  1. Jul 8, 2007 #1

    A block (mass = 2.4 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1.5 x 10-3 kg·m2), as the figure shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.032 m during the block's descent.
    Find (a) the angular acceleration of the pulley and (b) the tension in the cord.

    2. Relevant equations

    Newtons Second law and Newtons second law of rotation
    F=ma and Torque=Ialpha

    3. The attempt at a solution
    I tried using this equation, but i get the wrong answer no matter what I do.
  2. jcsd
  3. Jul 8, 2007 #2
    i think it must be


    because the tension is opposite to the gravitational pull......
    and where is the pic???
  4. Jul 8, 2007 #3
    ma=mg-T => a=g-T/m (1)
    also, a=alpha*radius of pulley and alpha= Torque/inertia = T*radius of pulley/inertia of pulley
    => a=radius^2*T/inertia (2)
    from 1 and 2, hopefully we can find the answer. I'm not sure why they mention the radius of the cord?
    Tell me if it works out, I didn't have time to actually solve it myself.
  5. Jul 8, 2007 #4
    Yes thankyou, that helped alot!!

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