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Two disks (moment of inertia)

  1. Nov 20, 2005 #1
    What is the final angular speed?

    I came up with 9.4 rad/s, which is correct.

    What fraction of the initial rotational kinetic energy is converted to heat in the process?

    I came up with .0588, which is also correct.

    A motor must restore the angular speed of the combination to wo in one revolution. What torque must the motor supply ?

    I used the equation [itex]\tau=I\alpha[/itex] here. I added each moment of inertia (for each disk individually):

    [tex]I=I_1+I_2=.02125[/tex]

    The above I got using the standard formula for the moment of inertia for a disk (1/2)mr2.

    Now I found the acceleration:

    [tex]10^2=9.4^2+2\alpha\left(2\pi\right)\implies\alpha=.926\text{rad}/\text{sec}^2[/tex]

    Then I multiplied the acceleration by the moment of inertia to come up with:

    [tex]\tau=.0197\text{Nm}[/tex]

    ...however the above is incorrect. Could someone please tell me why? I have gone through my steps many times, meaning that the only way this could be wrong is if I took the wrong steps; my arithmetic is fine. I'd appreciate any input on this.

    Thank you very much.
     
  2. jcsd
  3. Nov 20, 2005 #2
    your procedure is correct ... your final solution for tau does have some round-off error in it from rounding the final omega from the 1st part of the problem ... 9.4 rad/s.

    If you don't round, you should get tau = .0913092394 Nm

    If this is an on-line problem that is machine graded, that could be the reason.
     
  4. Nov 20, 2005 #3
    Thank you very much. It's nice to know my problem was in rounding digits rather than my procedure (it is web-based hw by the way).

    Thanks again.
     
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