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Rotational motion of a computer hard disk

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A computer hard disk starts from rest, then speeds up with an angular acceleration of 198 rad/s2 until it reaches its final angular speed of 7683 rpm. How many revolutions has the disk made 10.7 s after it starts up?

    2. Relevant equations

    3. The attempt at a solution
    when using the first equation i got t=804.561878/198=4.06
    then i plugged it into the second equation which would be delta0=0*4.06+(.5)(198)(6.63655)^2

    but it says that its wrong i dont understand what im doing wrong
  2. jcsd
  3. Oct 10, 2008 #2


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    Too much precision?
  4. Oct 10, 2008 #3


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    Up to speed means 7683 RPM = 128.05 rps
    In radians that's 804.53
    Divide that by 198 and that equals the 4.06 s
    How many radians is that? 1/2*198*(4.06)2 = 1631.88
    Now how many more at max speed.
    10.7 - 4.06 = 6.64 times 804.53 = 5342.01
    Add the two = 6973.96
  5. Oct 10, 2008 #4
    im sry for asking this but y did u times 6.64 by 804.53?? and then add the two answer
  6. Oct 10, 2008 #5


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    You have an interval of 10.7 seconds. The first 4.06 seconds was accelerating to max speed. So you need to calculate the number of radians it took to do that.

    But then it kept on spinning at the max speed didn't it? And that 6.64 seconds is how long it spun at max speed.

    So ... you calculate the number of additional radians until the time budget is expired. The two together is the total number of radians.

    Remember they were asking for revolutions so you still have that conversion.
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