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Problems w/ rotational dynamics and angular momentum

  1. Nov 14, 2005 #1
    1) A centrifuge rotor rotating at 10,300 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 m-N. If the mass of the rotor is 4.8 kg and it can be approximated as a solid cylinder of radius .0710 m, through how many revolutions will the rotor turn before coming to rest, and how long will it take? correct answer: 993 rev, 10.9 s

    So the first thing we are trying to find is rotational position (O) with a dash through it. And what we know is:

    intial w = 10,300 rpm, bought to rest by 1.20 m-N torque, m = 4.8 kg, r = .0710 m & we know the moment of inertia equation of a solid cylinder which is 1/2MR^2

    My thoughts we to find the moment of Inertia, then use the T = I(ang)a
    so I = .0121, then use the torgue equation(1.20 mN/.0121) and ang. acc. = 99.2 m/s^2
    then I could use rotational kinematics by: w2 = w(intial)^2 +2a(O)

    10,300 rpm^2 = 0 + 2(99.2m/s)(O)
    171.7 rps^2 = 198.4 m/s (O)

    O = 149 revs... not the correct answer, so assuming I got that wrong I cannot get the time...
    -------------------------
    2) A person stands, hands at his side, on a platform that is rotating at a rate of 1.30 rev/s. If he raises his arms to a horizontial position the speed of rotation decreases to 0.80 rev/s. (a) Why? (b)By what factor has the moment of inertia changed? (b) 1.6

    Okay, so I know part a which is because his rotational inertia increases. Part b is the problem...
     
  2. jcsd
  3. Nov 14, 2005 #2

    siddharth

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    In SI, angular veloity is in radians per second. Whereas in the question, it's given as revolutions per minute.

    For the second one, is there an external torque? So what does that tell you?
     
  4. Nov 15, 2005 #3

    FredGarvin

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    I'm not sure I agree with the answer for part B of question 2. Look at conservation of angular momentum.
     
  5. Nov 15, 2005 #4
    so I use this for nuber 2, but in rad/s?

    10,300 rpm^2 = 0 + 2(99.2m/s)(O)
    1078 rad/s^2 = 198.4 m/s
    O = 5.43 and then convert that into revs? that does not work either.
     
  6. Nov 15, 2005 #5
    As for # 2, I am stil clueless as to where the 1.6 is coming from.
     
  7. Nov 15, 2005 #6

    FredGarvin

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    You didn't square the [tex]\omega[/tex] term before you divided by 198.4.

    For the last one, since it is not mentioned, the system is a closed system without outside forces thus conservation of angular momentum should hold true. If you look at it like that (which I think is justified) you get a factor of 1.4, not 1.6.
     
    Last edited: Nov 15, 2005
  8. Nov 16, 2005 #7
    Okay so if I sq it.

    1078 rad/s^2 = 198.4 m/s
    1162084 rads/s = 198.4 m/s
    O = 5857 then to turn it into revolutions I divided by 2pie, correct?

    and I get 932.6 = 933 but that is still off by 60...

    O = 149 revs... not the correct answer, so assuming I got that wrong I cannot get the time...
     
  9. Nov 16, 2005 #8

    FredGarvin

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    I too get θ = 5880 = 936 rev. I can't see anything wrong with the process. Perhaps there is another error in the answers?

    Also, you can calculate the time before you calculate the rotations via [tex]\Sigma T = I \alpha[/tex] to calculate [tex]\alpha[/tex]. From there you can calculate the time via [tex]\omega = \omega_o + \alpha t[/tex]. You do not need the displacement to calculate t.
     
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