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Rotational motion -centrifuge!

  1. Mar 16, 2010 #1
    1.A centrifuge rotor rotating at 5000 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.60 m·N. If the mass of the rotor is 4.70 kg and it can be approximated as a solid cylinder of radius 0.0780 m, through how many revolutions will the rotor turn before coming to rest?



    2. Relevant equations Change rpms to rad/sec, Theta = Omega (initial)*time + 1/2*alpha*(t^2).. alpha*1/2*MR^2= Torque



    3. I get 1950 revolutions once i switch back to revolutions by mult by 180/pi..

    Is it my physics or my math? help please!
     
  2. jcsd
  3. Mar 17, 2010 #2

    ideasrule

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    Can you show your work so we can find the problem?
     
  4. Mar 17, 2010 #3
    1.6=1/2(4.7)(0.078)^2*alpha
    alpha=111.908 rad/sec^2..this is a negative quantity

    Omega initial=5000 rpm *pi/180=87.266 rad/sec

    Omega final=0= 87.266^2 +2(-111.908)(Theta)

    Theta= 34.02 rad*180/pi = 1949.51 revolutions
     
  5. Mar 17, 2010 #4
    I dont know where Im wrong..
     
  6. Mar 17, 2010 #5

    ideasrule

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    That is not how you convert from rpm to rad/sec. rpm means revolutions per second. How many radians is in a revolution? How about seconds in a minute? Make the appropriate conversion using those values.
     
  7. Mar 17, 2010 #6
    ah! thankyou
     
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