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Speed of a pendulum at lowest point

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data

    I am told to find the speed of a pendulum bob at the lowest point of the swing where the initial angle of displacement is θ=5.7° and the period of the pendulum is 2.5 seconds.

    2. Relevant equations

    K.E.=.5mv2, T=2∏√(L/g), P.E.=mgh, h=L-Lcosθ=L(1-cosθ)

    3. The attempt at a solution

    So I found the length of the pendulum first by plugging in 9.81 m/s2 for g and 2.5 s for T into the equation T=2∏√(L/g) to get 1.552 m.

    Next, we know that the height displacement in a pendulum bob is equal to L-Lcosθ, so I plug this into P.E.=mgh, and also from conservation of energy we know P.E. can be set as .5mv2, so the final equation I used is:
    v=√(2gL(1-cosθ))=√(2(9.81)(1.553(1-cos(5.7)))) which got me the answer v=2.244 m/s.

    However the answer is apparently v=0.388 m/s. My teacher is kind of notorious for being wrong a lot of the time, so if somebody could let me know which answer is right and why that would be awesome, thanks.
     
  2. jcsd
  3. Apr 14, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    Did you set your calculator to degrees (DEG) to calculate sin(5.7°)?
     
  4. Apr 14, 2013 #3
    Wow I feel stupid. Got the answer now, thanks.
     
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