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Two Metal Disks

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Two metal disks one with radius R1 = 2.50cm and mass M1=0.80kg and the other with radius R2=5.00cm and mass M2=1.60kg are welded together and mounted on a frictionless axis through their common center.
    a) a light string is wrapped around the edge of the smaller disk, and a 1.50kg block is suspended from the free end of the string. What is the magnitude of the downward acceleration of the block after it is released?
    b) repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk. In which case is the acceleration of the block greater?


    2. Relevant equations

    [itex]
    \tau=FR=I\alpha
    [/itex]

    3. The attempt at a solution
    a) My approach is that I will first try to find the angular acceleration of the disks

    F= the weight of the block, R=R1 and I=total inertia of the disks and alpha=angular acceleration of the disks. So...

    [itex]
    mgR_1=\frac{1}{2}\left(M_1R_1^2+M_2R_2^2\right)\alpha
    [/itex]

    plugging in the numbers, i got [itex]\alpha=163 rad s^{-2}[/itex]

    using the equation [itex]a_{tan}=R\alpha[/itex], I got a=4.08m/s^2 which seemed like a sensible answer, but not the correct answer.

    The correct answer is a=2.88m/s^2.

    Did I miss something? Or did I do something wrong??

    b) when I tried the same steps but this time using R2, I got a=16.3 m/s^2 which is totally nonsense. O.O||

    What did I do wrong? Please help.
     
  2. jcsd
  3. Oct 10, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    The mass is not falling freely. So the torque due to suspended mass is not equal to mgR1.
    If T is the tension in the string then
    ma = mg - T
    Now write down the expression torque and find the acceleration using the above equation.
     
  4. Oct 10, 2009 #3
    ah... now I get it! Thanks!
     
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