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Pendulum and velocity

  1. Apr 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A mass of 200 g is attached to the end of a cord 1.20 m long. The cord is hung from a horizontal beam and the mass is free to swing making a pendulum. The mass is released from rest at an angle of 60° from vertical. What will be the velocity of the mass at an angle of 30°, and at an angle of 0°?


    2. Relevant equations

    F_c = m(v²/r)


    3. The attempt at a solution

    Untitled7.png

    I figured that since it seems to be traveling in a half circle type of movement...I could use the fact that the length of the string would be the radius...so 1.20m

    after that I did

    ƩF_y = ma_y
    mgcosθ = m(v²/r)
    the m's cancel so
    gcosθ = v²/r
    v = √(rgcosθ)

    And I figured that I could use that same equation to solve for each of the angles...i.e

    = √((1.20(9.8cos60))
    = √((1.20(9.8cos30))
    = √((1.20(9.8cos0))

    Would this be correct? Or am I totally off here?
     
  2. jcsd
  3. Apr 11, 2013 #2
    you are on the right track but the angles marked on your diagram are wrong !!!
    You need to sort that out to make sense of your equations
     
  4. Apr 11, 2013 #3
    Untitled7.png

    Right, sorry about that I'm just so used to doing it "from horizontal"
     
  5. Apr 11, 2013 #4
    You are on the right track.
    A good check is to look at the extremes....when angle = 0, Cos0 = 1 and the velocity should be max....which it is !!
     
  6. Apr 11, 2013 #5

    SammyS

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    You're quite far off.

    How do you get that mgcos(θ) is equal to the centripetal force? It's not.


    mgcos(θ) = m v2/r can't possibly work, because for instance, at the instant that the mass is released, the velocity is zero, but θ = 60° .

    Use conservation of energy.
     
  7. Apr 11, 2013 #6
    In these calculations it is better if you can use angular frequency (ω)
    You will find the equations in SHM written as
    displacement x = ASin(ωt)
    velocity v = dx/dt = ωACos(ωt)
    acceleration =dv/dt = -ω^2.ASin(ωt)

    dont worry about this if you have not met it, you have done the right thing
     
  8. Apr 11, 2013 #7
    Right, makes sense...the lowest point here would be the max velocity. Just like if the angle was 90° the velocity would = 0.

    So okay then my equations would be correct?

    = √((1.20(9.81cos60)) = 5.886m/s
    = √((1.20(9.81cos30)) = 10.19m/s
    = √((1.20(9.81cos0)) = 11.77m/s ( which would be the max velocity )

    Look about right?
     
  9. Apr 11, 2013 #8
    Oh..then they are not correct.

    Well okay...assuming that this system is frictionless then W = 0

    It goes from Gravitational P.E to Kinetic Energy so

    mgh = 1/2mv²

    m's cancel so

    gh = 1/2v²

    rearrange for v

    V = √(2gh)

    and h would be 1.20cosθ here?
     
  10. Apr 11, 2013 #9
    How do you get that mgcos(θ) is equal to the centripetal force? It's not.


    mgcos(θ) = m v2/r can't possibly work, because for instance, at the instant that the mass is released, the velocity is zero, but θ = 60° .


    The angle θ = 60 for the pendulum diagram corresponds to (ωt) = 90 (the release point) in the equivalent circular motion and the amplitude A = LSinθ, ie A=LSin(ωt)

    mgCosθ = mv^2/r = centripetal force is correct.
    mgCosθ = ma where a = centripetal acceleration (v^2/r)....r = length of pendulum
     
    Last edited: Apr 11, 2013
  11. Apr 11, 2013 #10
    While I'm not 100% on what's happening here (mainly because we haven't met the use of those equations....we are in fact going over conservation of energy in class.......and what SammyS said about the release point being at 60° where the velocity = 0...then

    mgcos(60) = m(v²/r)
    some number = 0 .....so I actually don't see how this would work.

    but then again I'm not sure how to add centripetal acceleration into that equation.

    I'm 100% not sure here lol so I'll wait until you guys can come to a consensus about this one while I read up on it.

    Thanks guys
     
  12. Apr 11, 2013 #11
    the source of the problem is that the angle you use to describe the pendulum is not the angle that occurs in SHM equations.
    Your 60 is 90 (or 0) in the SHM equation
     
  13. Apr 11, 2013 #12
    I think that your use of SHM stands for harmonic motion? (simple harmonic motion I think it was?)

    While we haven't reached that yet in lecture...I remember reading about it....and if I remember correctly....what you're saying is that the fact that there is no angular displacement at 60° then it would = 0? Is that correct?
     
  14. Apr 11, 2013 #13

    SammyS

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    Are you replying to me, since you quoted from my post ? (There is a QUOTE feature that is provided.)
     
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