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Conservation of energy and string pulley problem

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider two masses of 3.3 kg and 7.4 kg
    connected by a string passing over a pulley
    having a moment of inertia 12 g · m2
    about its axis of rotation, as in the figure below. The
    string does not slip on the pulley, and the
    system is released from rest. The radius of
    the pulley is 0.35 m.

    Find the linear speed of the masses after
    the 7.4 kg mass descends through a distance
    21 cm. Assume mechanical energy is conserved during the motion. The acceleration of
    gravity is 9.8 m/s^2.

    Answer in units of m/s

    2. Relevant equations

    delta PE = KE
    KE = K(translational) + K(rotational)

    3. The attempt at a solution

    Found that change in potential energy should be equal to change in potential energy of the heavier mass (where PE is lost) subtracted by the change in potential energy of the lighter mass (where some PE is gained).

    Therefore: 7.4*g*(21/100) - 3.3 * g * (21/100) = delta PE

    I then set total change in KE to the delta PE. I determined rotational KE to be 1/2 I * omega^2

    where I is given to me and omega is (v/r)^2 and r is given to me.

    K translational = 1/2 * Mtotal * v^2. I factored out V^2, and set that equal to delta PE / rest of that mess

    So in the end: v = sqrt ( delta PE / (6/r^2 + 1/2M)).

    However, it wasn't the right answer :( so any ideas?
     
  2. jcsd
  3. Nov 3, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    In what units is the moment of inertia given?
     
  4. Nov 3, 2011 #3
    wow I see the moment of inertia is given in grams. Thank you so much that would explain why I failed XD
     
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