Kinematics and dynamic circular motion of conical pendulum

diazdaiz

1. Homework Statement 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. Homework Equations
V=ωR
Fcentripetal = $\frac {MV^2} {R}$
Fgravity = MG
phytagoras
basic trigonometry

3. The Attempt at a Solution 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)

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Replusz

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.

diazdaiz

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.
ok, thank you very much

"Kinematics and dynamic circular motion of conical pendulum"

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