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Kinetics of a rigid body (conservation of energy)

  1. Jul 17, 2015 #1
    1. The problem statement, all variables and given/known data
    eZTS4A6.jpg



    2. Relevant equations

    (1) (pink) Is it vertical displacement ?
    * 4 - 2 (unstretched lenght of the spring) ?

    (2) (blue) What does it mean ?
    * I was thinking that maybe they used The Pythagorean theorem (62+42)=7,2 ≅ 7) but I'm not sure. Besides I don't understand why they used it.

    3. Attempt

    My reasoning: In general I think they are using this formula,
    upload_2015-7-17_13-20-26.png
    where in that case the left side is the initial state - the right side - the final state of the spring.
     

    Attached Files:

  2. jcsd
  3. Jul 17, 2015 #2
    I recommend setting it up carefully in terms of conservation of energy, expressing T and V in terms of all the relevant quantities (leaving as symbols), making simplifications, solving for the unknown (with symbols), and plugging in numbers at the end.
     
  4. Jul 17, 2015 #3

    haruspex

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    The 4-2 and 7-2 are both calculations of spring extension. In each case, 2 is the relaxed length. At the horizontal position, the total length is 4, so the extension is 4-2. After rotating 30 degrees, the total length is 4+6 sin(30)=7.
    I don't like the way they divide the mass by 32.2 in the KE terms. It seems more natural to me to multiply by 32.2 in the other terms - it comes to the same thing, of course. Maybe their way is standard in this non-metric system. I have no background in it.
     
  5. Jul 17, 2015 #4
    The English (Imperial) unit of mass is slugs. One obtains mass by dividing weight (lbs) by the acceleration of gravity (32.2 ft/s/s) to obtain the mass in slugs.

    I thought ballistics was the last holdout still using English units for real physics.

    https://en.wikipedia.org/wiki/English_Engineering_units
     
  6. Jul 17, 2015 #5

    haruspex

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    Ok, thanks.
     
  7. Jul 20, 2015 #6
    Thank you for help.

    Dr. Courtney - is this equation correct ?

    X0yeXaC.png
     
  8. Jul 20, 2015 #7
    Is the only potential energy due to the spring?
     
  9. Jul 20, 2015 #8
    If a spring is not stretched or compressed, then there is no elastic potential energy stored in it.

    However in this case at the beggining the spring was stretched 2 metres (4-2=2), hence 1/2*k*s1^2

    At the end, the spring was stretched 5 meters (7-2=5), hence 1/2*k*s2^2

    'W' stands for: work of a weight -> WΔy= 50 * (1.5)
     
  10. Jul 20, 2015 #9
    Gravity?
     
  11. Jul 20, 2015 #10
    I think the only potential energy is due to the spring. If the spring lies the ground then mgh=0
     
  12. Jul 20, 2015 #11

    haruspex

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    The massive bar is rotating in a vertical plane. Are you saying its centre of gravity does not change in height?
     
  13. Jul 22, 2015 #12
    Ok, it does. Thanks.
     
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