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Energy transforations

  1. Oct 20, 2007 #1
    Energy transforations!!

    A student swings on a swing so that his centre of mass is located 2.2 m from the rest point where the rope is attached to the rail. If he swings so that his maximum amplitude causes the rope to make an angle of 49° with the vertical, calculate the child's maximum speed during the swing.


    I dont know where to start!!
     
  2. jcsd
  3. Oct 20, 2007 #2
    ok...i've got a slight idea..we have to take the motion of the spring as a pendulum..
     
  4. Oct 20, 2007 #3

    learningphysics

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    use conservation of energy. gravitational potential energy... kinetic energy.
     
  5. Oct 20, 2007 #4
    ok..soo at the equlibrium position, we use total energy=mgh where h is 2.2m, and at the new postition the total energy is 0.5mv^2 yes..?
     
  6. Oct 20, 2007 #5

    learningphysics

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    almost there... 2.2m is not your h. 2.2m is the length of the rope (not exactly but it is the distance from the joint to the center of mass of the child)...

    How high above the bottom does the child's center of mass reach? use 2.2m, 49 degrees and trig. that would be your h.
     
  7. Oct 20, 2007 #6
    ok..so i use tan49= h/2.2, hence i can find h..but what abt the mass ??..and wait should'nt the energy at the equilibrium position be the kinetic energy, since h is 0..?
     
  8. Oct 20, 2007 #7
    so when i take the total energy at the equilibrium position as the kinetic energy, and the total energy at the other position as the potential energy, and solve them, i got v as 7.04m/sec..is tht ok..?
     
  9. Oct 20, 2007 #8

    learningphysics

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  10. Oct 20, 2007 #9
    ok..soo by doing that i got v as 3.86m/s
     
  11. Oct 20, 2007 #10

    learningphysics

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    looks good. I get 3.851m/s though... it's just the rounding.
     
  12. Oct 21, 2007 #11
    If [tex] l [/tex] is the length of the rope then, the height will be [tex] h = l(1 - \cos{\theta} ) [/tex]
    The velocity of the boy is then simply [tex] \sqrt{2gh} [/tex] at the equilibrium position. At equilibrium position the velocity of a pendulum is maximum, so that's what you're looking for.
    You get the equation by using simple conservation laws. (That of the potential energy at the height and the kinetic energy at the bottom.
    I know you've got the answer, but I'm just reiterating the facts so that it becomes clear why you do what you do.
    Moreover try to use variables rather than direct numerical values, you'll miss the physics involved, and it might get complicated unnecessarily.
     
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