# Kinematics and dynamic circular motion of conical pendulum

1. May 15, 2016

### diazdaiz

1. The problem statement, all variables and given/known data

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

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

2. May 15, 2016

### 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.

3. May 15, 2016

### diazdaiz

ok, thank you very much