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Rotating Disks - How to calculate final KE?

  1. Aug 9, 2006 #1
    Hi, I have worked through this problem for so long that I cannot see how I could be getting it wrong. I feel like I am doing it correctly, but I don't get the right answer.

    A disk of mass M1 = 350 g and radius R1 = 10 cm rotates about its symmetry axis at finitial = 154 rpm. A second disk of mass M2 = 260 g and radius R2 = 7 cm, initially not rotating, is dropped on top of the first. Frictional forces act to bring the two disks to a common rotational speed ffinal.
    a) What is ffinal? Please give your answer in units of rpm, but do not enter the units.
    ff= 112.9 rpm​

    b) In the process, how much kinetic energy is lost due to friction?
    |KElost| = ??? J ​

    I figure that the change in KE is equal to .5*(Ii*wi^2)-.5*(If*wf^2). KEi is easily calculated and is 0.22756 J.

    Is it fair to assume that the If in KEf is the sum of the I's for both disks, because they are now spinning together? I did that, came up with I(sum)=0.002387, which gives a If of 0.11683.

    KEi - KEf = change in KE
    0.22756-0.11683 = 0.06073 - which is wrong.

    Should I be incorporating translational KE into this? I tried that and also go the wrong answer.

    Thank you for your insight.
  2. jcsd
  3. Aug 9, 2006 #2


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    Staff Emeritus
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    Here is your problem. [itex]0.22756-0.11683\neq0.06073[/itex]. Looks like you pushed the wrong buttons on your calculator :smile:
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