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Kinetic energy and Angular momentum

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Five identical particles of mass m = 0.30 kg are mounted at equal intervals on a thin rod of L = 0.87 m and mass M = 2.0 kg, with one mass at each end of the rod. If the system is rotated with angular velocity ω = 50 rev/min about an axis perpendicular to the rod through one of the end masses, determine the following.

    (a) the kinetic energy of the system

    (b) the angular momentum of the system


    2. The attempt at a solution
    SOLVED
    (a) K=.5(Itotal)w^2
    K=.5(1/3L2*M + 30/16L2*m)w2
    K=12.75312242kg·(m/s)2
    (b)Angular Momentum=Itotal*w
    Angular Momentum=4.87133934kg·m2/s
     
    Last edited: Mar 17, 2009
  2. jcsd
  3. Mar 17, 2009 #2

    LowlyPion

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    They want you to develop the I for this configuration.

    If there are 5 evenly spaced, then you have to determine the distances (L/4 is the separation) and then develop the sum of the moments of inertia of the parts. The rod and the weights. However this is a little trickier. The weight in the middle needs to be properly accounted for.

    So you can use the I of a rod rotated about it's end and then the 4 weights evenly spaced at L/4 along its length.
     
  4. Mar 17, 2009 #3
    Well what you're describing pretty much sounds like what I did, Find the I of the rod which is just (1/3)Md^2 where d = .87m. Then find the Inertia of each particle with a varying L which would come out as L1=0, L2=.2175, L3=.435,L4=.6525,L5=.87. so the general form for the Inertia of the particles is .5m*(Ln)^2 (n being 1-5). then just add all Irod and Iparts. multiply by (w^2)/2.
     
    Last edited: Mar 17, 2009
  5. Mar 17, 2009 #4

    LowlyPion

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    From the end I get (1/3L2*M + 30/16L2*m) for I
     
  6. Mar 17, 2009 #5
    what do you mean by "from the end"
     
  7. Mar 17, 2009 #6

    LowlyPion

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    From the problem right? Don't you need I relative to one end?
     
  8. Mar 17, 2009 #7
    yeah, relative to the end its rotating about. but is that I equation you gave supposed to be the total I? more specifically is the (30/16L2*m) part the I for all the particles summed together? and if it was all the particles added together what would L be?
     
  9. Mar 17, 2009 #8

    LowlyPion

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    m is the mass of the little weights.

    (L/4)2 + (L/2)2 + (3L/4)2 + L2 = 30/16 L2

    1/16 + 4/16 + 9/16 + 16/16 = 30/16 right?
     
  10. Mar 17, 2009 #9
    ah, I see. That makes so much more sense now XD. so Itotal=(1/3L2*M + 30/16L2*m), so to get K it would be (1/2)(Itotal)w2 and in this case w=50rev/min

    so Itotal=(1/3(.87m)2*2kg + 30/16(.87m)2*.3kg)=.93035625kg*m2
    K=(1/2)(.93035625kg*m2)(5.235987756/s))2=12.75312242 kg*(m2/s2)
    OK got it, thank you SO MUCH Pion.
     
    Last edited: Mar 17, 2009
  11. Mar 17, 2009 #10

    LowlyPion

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    There you go. You want radians/sec.
     
  12. Mar 17, 2009 #11

    LowlyPion

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    Much better.

    Good luck.
     
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