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Homework Help: The coservation of energy principle

  1. Nov 3, 2007 #1
    1. The problem statement, all variables and given/known data

    I had just finished solving 9 energy problems and I'm stuck with this hard question:

    A simple pendulum, which consisits of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates by swinging back and forth in a vertical plane. If the string is 2.00 meters long and makes an initial angle of 30.0 (degrees) with the vertical, calculate the speed of the particle(a) at the lowest point in its trajectory (b) when the angle is 15.0(degrees).

    I didn't understand the whole idea of this question ( I mean I can't draw it in my mind)

    so plz i'll appreciate any help


    2. Relevant equations
    The law of conservation of energy:
    M.E.(f) = M.E.(i)

    K.E.(f) + P.E.(f) = K.E.(i) + P.E.(i)

    K.E.( kinetic enrgy) = 1/2 m(V)^2
    P.E.(gravetational potential energy) = mgh or (P.E.=mgy)

    I think there is no elastic potential energey
    3. The attempt at a solution

    I didn't understand the idea of question
  2. jcsd
  3. Nov 3, 2007 #2
    I'm online for any discussion
  4. Nov 3, 2007 #3
    Choose a reference line to calculate PE, and equate total energy initially and finally(at lowest point in first case).
  5. Nov 3, 2007 #4
    ok but what is the initial hight

    How can i calculate the hight
    Last edited: Nov 3, 2007
  6. Nov 3, 2007 #5
    height - distance of the point from reference line.
  7. Nov 3, 2007 #6
    I did that: my reference is the lowest point of the trajectory

    K.E.(f) + P.E.(f) = K.E.(i) + P.E.(i)

    (1/2 m(V)^2 + mgh)(f) = (1/2 m(V)^2 + mgh)(i)

    1/2 m(V(f))^2 = mgh(i)

    how can i find initial hieght h(i)
  8. Nov 3, 2007 #7
    can we use the string's length?
  9. Nov 3, 2007 #8
    Nope. Use trigonometry. Initially string is at 30 deg with vertical, can you calculate height using this(remember, height is of bob)?
  10. Nov 3, 2007 #9
    ok is this the answer of a)

    K.E.(f) + P.E.(f) = K.E.(i) + P.E.(i)

    (1/2 m(V)^2 + mgh)(f) = (1/2 m(V)^2 + mgh)(i)

    1/2 m(V(f))^2 = mgh(i)

    1/2 m(V(f))^2 = mg(5sin30)

    (V(f))^2 = 2g2.5

    V = 7m/s
  11. Nov 3, 2007 #10
    right or wrong?

    plz tell me

    I'll be very thankful
  12. Nov 3, 2007 #11
    Well, how did you get h(i) = 5 sin30 (remember your ref line is at bottom of trajectory, so h(i) is just the vertical distance of bob at initial time from ref line)
  13. Nov 3, 2007 #12
    I used trigonemotry to find it as you say
  14. Nov 3, 2007 #13
    try it i think my answer may be correct
  15. Nov 3, 2007 #14
    h(i) = l (1 - cos [tex]\theta[/tex][tex]_{i}[/tex])
  16. Nov 3, 2007 #15
    Convince yourself about this or ask.
  17. Nov 3, 2007 #16
    ok now I can solve the question using this

    but i still don't know from where did you bring (1-cos[tex]\theta[/tex])
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