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Dynamics: max horizontal swing of a pendulum

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data

    In an ore-mixing operation, a bucket full of ore is suspended from a traveling crane which moves along a stationary bridge. The crane is traveling at a speed of 3 m/s when it is brought to a sudden stop. Determine the maximum horizontal distance through which the bucket will swing. (arm length 10m)


    2. Relevant equations

    KE = 1/2mv2
    PE = mgz
    Etot = KE + PE + U

    3. The attempt at a solution

    Since the bucket is traveling with a speed of 3 m/s and has no height E1 = 1/2mv2 and at the highest point the bucket will travel v = 0 m/s E2 = mgz

    Therefore since E1 = E2 (law of conservation of energy)
    1/2mv2 = mgz
    1/2v2 = gz
    1/2(3 m/s)2 = (9.81 m/s2)z
    z = 0.4587 m

    I'm not even sure if I did this the way my professor taught us, I'm just using thermo energy balance equations to be honest. I don't know if there is a way to make a triangle and solve for the x component, or just a complete different way to do this.
     
  2. jcsd
  3. Feb 23, 2012 #2
    Another way you can try to conserve momentum
     
  4. Feb 23, 2012 #3
    I used a triangle to find out that the 10 m arm length minus the 0.4587 m height increase equals a triangle with the vertical leg being 9.54 m and the hypotenuse of 10 m. I then did sin-1(9.54 m/10 m) = 72.55°

    Then I did 10cos(72.55°) = 2.9987 m

    I know this is the right answer but I don't know if there is another way to find this answer.
     
  5. Feb 23, 2012 #4
    Are masses given ?
     
  6. Feb 23, 2012 #5
    No, no masses are given
     
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