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Torque By a Mass on a Disk

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data
    An 8.0-cm radius disk with a rotational inertia of 0.12 kg m2 is free to rotate on a horizontal axis. A string is fastened to the surface of the disk and a 10-kg mass hangs from the other end. The mass is raised by using a crank to apply a 9.0-N m torque to the disk. The acceleration of the mass is:


    2. Relevant equations
    torque = I(ang. accel)


    3. The attempt at a solution
    I have no clue. I tried solving for the angular acceleration, but I don't know how to connect the acceleration of the mass to the acceleration of the disk.
     
  2. jcsd
  3. Nov 19, 2007 #2

    Astronuc

    User Avatar

    Staff: Mentor

    See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/mi#cmi

    and

    http://hyperphysics.phy-astr.gsu.edu/hbase/mi - general formulae

    http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#rq - rotational quantities.


    Given: 8.0-cm radius disk, so r = 0.08 m
    rotational inertia of 0.12 kg-m2, so I = 0.12 kg-m2

    Find the appropriate expression for I of a disk.

    The disk is rotating on a horizontal axis. The suspended mass has a weight pulling it down, and the crank applies a torque to the disk and suspended weight at moment r.

    What is the net moment on the suspended mass?
     
  4. Nov 19, 2007 #3
    Is the moment arm of the force assumed to be r? From the question, I thought the string was attached to an arbitrary point on the disc, and a crank was spinning the mass on the other end of the string.

    I don't know how to calculate the net moment. It seems that the net moment would depend on the length of the string.
     
    Last edited: Nov 19, 2007
  5. Nov 19, 2007 #4
    The linear acceleration of the mass should be the same as that of the disk. If we just find the linear acceleration of the disk with net torque 9 Nm, we get the answer.

    I think the answer is 6 m/s^2
     
  6. Nov 19, 2007 #5
    The general idea how to solve these problems is the following:

    Apply Newtons Second Law in Linear Motion --> Sum F = m*a
    Apply Newtons Second Law for Rotational Motion --> Sum Torque = I * alpha

    In any closed system there is the relationship that a = r*alpha, which means that the linear acceleration is the same for both the mass and the disk.

    I believe that if you use those two steps it will be easier to understand whether your answer is correct or not.

    I suspect that your answer is incorrect there should be torque from both the force that the gravity force of the mass(each in opposite directions)
     
  7. Nov 19, 2007 #6
    Anyone have a possible solution? I feel this problem is impossible. The acceleration of the mass would undoubtedly be dependent on the length of string.
     
  8. Nov 19, 2007 #7
    Like i said above:
    For the mass hanging on the string:
    Sum F = m*a <=> T- mg = ma --> T = ma + mg

    For the disk:
    Sum Torque = I*alpha --> F*r - T*r = I * alpha or F*r - T*r = I*a/r

    Substituting the value of the Tension we have above, we get:

    F*r - (ma+mg)*r = I*a/r ==> a = (F*r - mgr)/((I/r) + m)

    Now as much as i would like to plug in values i am usually satisfied with the variable solution only. I hope this helps!
     
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