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Kinematics and dynamic circular motion of conical pendulum

  1. May 15, 2016 #1
    1. The problem statement, all variables and given/known data
    one.png

    find the period with only using L (for the long of the rope), R (for the radius), M (for the mass), and G (for the gravity)

    2. Relevant equations
    V=ωR
    Fcentripetal = ##\frac {MV^2} {R}##
    Fgravity = MG
    phytagoras
    basic trigonometry

    3. The attempt at a solution
    two.png

    i have tried to do it this way
    ##x=\sqrt {L^2 - R^2}##
    ##F_1=F_2##
    ##MG Cos (θ) = \frac {MV^2} {R} Cos (90-θ) ##
    ##\frac {MGR} {L} = \frac{MV^2 \sqrt {L^2 - R^2}} {RL}##
    ##\frac {MGR} {L} = \frac{Mω^2 R^2 \sqrt {L^2 - R^2}} {RL}##
    ##\frac {MGR} {L} = \frac{M4π^2 R^2 \sqrt {L^2 - R^2}} {RLT^2}##
    ##T^2 = \frac{4π^2 \sqrt {L^2 - R^2}} {G}##
    ##T = \sqrt {\frac{4π^2 \sqrt {L^2 - R^2}} {G}}##
    am i right?
    some of my friend have i different answer from me, actually, i dont really know where is the centripetal force direction

    can someone explain me what is centripetal force actually with answering this question
    (sorry for bad english)
     
  2. jcsd
  3. May 15, 2016 #2
    Hi,
    I think your solution is correct.
    Centripetal acceleration is acceleration which makes something 'go in a circle' - not straight forward. It points toward the centre of the circle.
    Multiplied with the mass, this gives the 'centripetal force'.
    Hope this helped.
     
  4. May 15, 2016 #3
    ok, thank you very much
     
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