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Homework Help: How to calculate velocity in this question

  1. May 31, 2010 #1
    1. The problem statement, all variables and given/known data

    a pendulum of mass 1.0kg and length 1.0m is swung from a height of 0.110m above its equilibrium position. Find the tension of the pendulum string at the bottom of its swing.

    Formula:
    T=mv^2/r+mg

    How do i get the V^2?
     
  2. jcsd
  3. Jun 1, 2010 #2
    I'm trying to find the tension...I don't have the # for T. do i just use v^2=Vi^2+2ad? and make velocity initial 0?
     
  4. Jun 1, 2010 #3
    no its periodic motion not projectile
     
  5. Jun 1, 2010 #4
    what is the # for T and u?
     
  6. Jun 1, 2010 #5
    do you mean coefficient by #????
    the velocity at the bottem is given by -wAsin(wt) and from x=Acos(wt) you know that wt =1 since velocity is max at the bottem
     
  7. Jun 1, 2010 #6
    yeah...what's the coefficient for T and u
     
  8. Jun 1, 2010 #7
    dont worry about what i said before it was for something else,,,, I get v at the bottom as 0.3445m/s, is the T in the formula up the top period or tension just asking because ive never seen it before.
     
  9. Jun 1, 2010 #8
    it's for tension
     
  10. Jun 1, 2010 #9
    do you get the right answer when you sub the v i found into your equation??? answer 9.93N
     
  11. Jun 1, 2010 #10
    yeah i got that answer (I don't know if it was right though since I don't have the answer key). How did you find V again? I have no idea what '-wAsin(wt) and from x=Acos(wt)' was.
     
  12. Jun 1, 2010 #11
    use energy conversion to get mgh = (1/2) mv^2. v is the speed at the bottom.
    and then T = mg + mv^2 / r
     
  13. Jun 1, 2010 #12
    that equation doesn't give me a velocity of 0.3445m/s.
    mgh=1/2mv^2
    (1kg)(9.8m/s^2)(1.1m)=1/2(1kg)(v^2)
    =4.64m/s

    Is it 4.64 m/s?
     
  14. Jun 1, 2010 #13
    t my procedure should yield exactly the same answer as using energy methods, T=2pisqrt(L/G) T=2.0 f= 0.498 w=3.13 v=wA v=0.244m/s , anyone know why that doesn't work?
     
  15. Jun 1, 2010 #14
    from energy methods v=1.469 but im not sure i agree with it since your acceleration (9.81) is for free fall but this is not the case for pendulums which is why our answer dont line up.

    there is a component of that acceleration in the x direction, im quietly confident that you cant use energy methods to solve this now,,,, more proof there is a horizontal displacement that looses energy. in conclusion you cant use energy methods!!!!!!!!!
     
  16. Jun 1, 2010 #15
    I have no idea, can someone clarify this please?
     
  17. Jun 1, 2010 #16

    rl.bhat

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    Homework Helper

    The above equation should be

    (1kg)(9.8m/s^2)(0.11m)=1/2(1kg)(v^2)
     
  18. Jun 1, 2010 #17
    Equations you need: note w is onega or angular velocity
    X(t)=Acos(wt)
    now you know that at equilibrium velocity is a maximum. so at this point A=x
    cos 0 =1 therefore wt=1
    T(period)=2pi*sqrt(L/g)
    T=2 , f=t^-1 f=0.498Hz , w =2pif, w=3.13
    v(t)=wAsin(wt) as said before, wt is 1
    vmax=wa
    vmax=3.13*0.11
    v=0.344m/s

    need more clarification??? ps as i said before you cant use energy methods
     
  19. Jun 1, 2010 #18
    rl.bhat do you think that that will give the correct velocity at the bottom??? ie do you think you can use energy methods for this?
     
  20. Jun 1, 2010 #19

    rl.bhat

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    Homework Helper

    In the simple pendulum, for small oscillations, the total energy remains the constant.
     
  21. Jun 1, 2010 #20
    The energy in the system would certainly remain constant. I dont understand how mgh=1/2mv^2 would possibly work though. the velocity that gives would be the velocity if it was dropped straight down with nothing attached. the velocity i calculated is significantly different than the one using energy and this oscillation would be called a "small oscillation" can you see anything wrong with what i did??
     
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