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Angular Momentum of a Uniform Rod

  1. Jul 3, 2015 #1
    << Mentors have notified the OP to show their Attempt at a Solution >>

    1. The problem statement, all variables and given/known data

    A uniform rod of length L1 = 1.5 m and mass M = 2.8 kg is supported by a hinge at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass m is supported by a thin string of length L2 = 1.1 m from the hinge. The particle sticks to the rod on contact. After the collision, θmax= 45°.

    a)Find m (in kg)

    b) How much energy is dissipated during the collision? (in J)

    10-55.gif


    2. Relevant equations

    I found the question with different numbers but when I plug in mine. It's still wrong.
    Screen Shot 2015-07-03 at 4.46.35 PM.png

    Screen Shot 2015-07-03 at 4.38.58 PM.png Screen Shot 2015-07-03 at 4.41.08 PM.png Screen Shot 2015-07-03 at 4.41.18 PM.png Screen Shot 2015-07-03 at 4.41.28 PM.png Screen Shot 2015-07-03 at 4.41.35 PM.png

    3. The attempt at a solution
     
    Last edited by a moderator: Jul 3, 2015
  2. jcsd
  3. Jul 3, 2015 #2

    SammyS

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    Hello Bob4321. Welcome to PF.

    Please show your attempt. What did you do in attempting to use that solution?
     
  4. Jul 4, 2015 #3
    Solution Attempt:

    Ok so for A:

    ω1=((1/3)(2.8)(1.52))/ ((1/3)(2.8)(1.52))+m(1.12)√3(9.81m/s2)/1.5= (4.429kg/s)/(2.1kg)+1.21m

    ½(4.429kg/s)2/(2.1kg)+1.21m I don't know what to do now. Where did they get the .2 from in their example?
     
    Last edited: Jul 4, 2015
  5. Jul 4, 2015 #4

    SammyS

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    What is (½)(1 - cos(θmax) ) ?

    It looks like you have an issue regarding "order of operations". The entire expression, (⅓)(2.8)(1.52)+m(1.12) should be in the denominator. It's clear that in the solution you're attempting to mimic, the mass, m, is in the denominator.


    In my opinion: This method of obtaining an answer for an exercise is not to be recommended. If you don't know what quantities are being used, and why they're used, simply getting an answer to work out numerically isn't of much use.

    Examining a worked out solution can indeed be very helpful, but don't just try to plug in some numbers into something that you don't understand.
     
  6. Jul 4, 2015 #5
    Ok I have figured out that the .2 is from 1-cos but I still keep getting the wrong answer! Could you solve it so I can see what you did?
     
    Last edited by a moderator: Jul 4, 2015
  7. Jul 4, 2015 #6
    For my problem the (1-cos45) is .48
     
  8. Jul 4, 2015 #7

    SammyS

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    That's not how we do things here at PF.

    0.48 Is incorrect. Were you using 45 radians rather than 45° ?
     
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