Kinematics and dynamic circular motion of conical pendulum

[diazdaiz]

Homework Statement

[/B]
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)

Homework Equations

V=ωR
F_centripetal = $\frac {MV^2} {R}$
F_gravity = MG
phytagoras
basic trigonometry

The Attempt at a Solution

[/B]
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 don't 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)

 

[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