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Homework Help: Trebuchet, max speed

  1. Aug 16, 2008 #1
    1. The problem statement, all variables and given/known data

    1. The problem statement, all variables and given/known data
    A war-wolf, or trebuchet, is a device used during the Middle Ages to throw rocks at castles and now sometimes used to fling pumpkins and pianos. A simple trebuchet is shown in Figure P8.77. Model it as a stiff rod of negligible mass 3.00 m long and joining particles of mass 60.0 kg and 0.120 kg at its ends. It can turn on a frictionless horizontal axle perpendicular to the rod and 14.0 cm from the particle of larger mass. The rod is released from rest in a horizontal orientation. Find the maximum speed that the object of smaller mass attains.
    Image: http://www.grabup.com/uploads/e7490c945283118e547daa00676d2d28.png
    2. The attempt at a solution

    don't even know where to start, any help is much appreciated
    (the answer is 24m/s but i don't know how to get that)
    Thanks
     
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Aug 16, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi Zukie91! Welcome to PF! :smile:

    Hint: use conservation of energy (KE + PE = constant). :smile:
     
  4. Aug 16, 2008 #3
    i noticed that there is another thread asking the same question which nobody replied to. On that thread, under relevant equations, KvnBushi listed this
    [tex]K = \frac{1}{2} I_{cm} w^2 + \frac{1}{2} M v_{cm}^2[/tex]
    is that actually relevant?
    ok i can't get that latex code or whatever it is to work properly, i'll just link the other thread
    https://www.physicsforums.com/showthread.php?t=200958
     
  5. Aug 16, 2008 #4

    tiny-tim

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    (LaTeX isn't working at the moment. :cry:)

    The Iw2 part isn't relevant.

    The mv2 part is. :smile:
     
  6. Aug 16, 2008 #5
    i appreciate you help btw,
    do i need to use torque?
    i'm sorry i'm just not getting this problem
    and how do i use pe if i don't have the height?
    oh wait, would the 14 cm be the height?
    and i made a typo on the answer, it should be 24.5 m/s not 24 m/s
     
    Last edited: Aug 16, 2008
  7. Aug 16, 2008 #6

    tiny-tim

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    No … just KE + PE = constant.
    sort-of … but 14cm is the length of the whole rod, and it isn't the end of the rod that's fixed, is it? :wink:
     
  8. Aug 16, 2008 #7
    it says the whole rod is 3.00 m long, and that the axel is 14.0 cm from the larger mass, which means its 2.86 m from the projectile right?
     
  9. Aug 16, 2008 #8

    tiny-tim

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    … oops!

    oops … I misread the question! :redface:

    Yes, you're right … the PE for the two masses will be based on .14 and 2.86 (at maximum height difference).

    Now how will you work out the KEs? :smile:
     
  10. Aug 16, 2008 #9
    so am i going to be using
    1/2mv^2i + mghi = 1/2mv^2f + mghf
    i'm not sure when i should use which mass, or if i should add/subtract them
     
  11. Aug 16, 2008 #10

    tiny-tim

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    ah … KE + PE = constant, so if you increase the KE from 0, you must decrease the PE … so it's minus. :smile:
     
  12. Aug 16, 2008 #11
    what is minus?
    and how does this only have to do w/ pe and ke, and nothing to do with rotation, if its rotating around an axle?
     
  13. Aug 16, 2008 #12

    tiny-tim

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    The gh has a minus in front of it: m(v2/2 - gh).

    And the question tells you to model the masses as particles, so all you need to know is their speeds. :smile:
     
  14. Aug 16, 2008 #13
    were you able to get the correct answer with this method, because i think i get what you are saying, and cannot get the problem right using any combination of m and h
     
  15. Aug 16, 2008 #14
    ok i don't think i'm really getting anywhere, would you mind setting up the eq, and i'll try to figure out what i was doing wrong from there?
     
  16. Aug 17, 2008 #15

    tiny-tim

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    Hi Zukie91! :smile:
    Been asleep … :zzz:

    I've checked the answer, and 24.5 is roughly correct.

    Hint: if the small mass has speed 24.5, then the large mass has speed 24.5 x 0.14/2.86 = 1.2

    Does that help? :smile:

    (if not, show us your calculations)
     
  17. Aug 17, 2008 #16
    still not getting it :( i'm closer but not right yet
    i'm doing
    deltaPE=deltaKE
    mgh1-mgh2=1/2Mv12+1/2mv22
    (60*9.8*.14) - (.12*9.8*2.86) + 30v12 + .06v22
    78.96=30v12 +.06v22
    then using the proportion v1/v2 = .14/2.87 i get that v2 = 20.5 v1
    substituting that in for v2
    i get 78.96=31.23v2
    which simplifies to v=1.59 which when divided by .14/2.86 is 32.48 which is still wrong :(
     
  18. Aug 17, 2008 #17

    tiny-tim

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    Hi Zukie91! :smile:

    Your 31.23 is wrong … you forgot to square. :frown:
     
  19. Aug 17, 2008 #18
    wait, what did i forget to square?
     
  20. Aug 17, 2008 #19

    tiny-tim

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  21. Aug 17, 2008 #20
    oh, duh, thank you so much. really appreciate all your help
     
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