# Homework Help: Circular Motion Derivation

1. Oct 12, 2008

### Epsillon

Hi for a lab I need to derivate the following formula.
T=√((4π^2 lm)/Mg)

I know Fnet= (mg/cos)sin

So so far I know Fnet= m x (4pi^2r)/T^2
and I know that r=lcos

but combining the above I got no where since I get an extra sin in there

Last edited: Oct 12, 2008
2. Oct 12, 2008

### Epsillon

Any help?

3. Oct 12, 2008

### tiny-tim

Welcome to PF!

Hi Epsillon! Welcome to PF!
No, r = … ?

4. Oct 12, 2008

### Epsillon

actually r=lsin(pheta)

5. Oct 12, 2008

### Epsillon

Ok so that was actually the mistake preventing me from solving the formula.

So I finally got it and I typed this up in words and posted it here incase someone needs it. If it doesent make sense I can perhaps attach the file.
Can someone verify that what I did is right?

Deriving the formula
F_net=ma_c=m((4π^2 r)/T^2 )
F_net= F_c=〖Ft〗_x=Ft×sinθ
Ft×sinθ= (mg/cosθ)sinθ =mgtanθ
Now one can combine the two calculated formulas for F_net
m((4π^2 r)/T^2 ) =(mg/cosθ)sinθ

r=lsinθ
∴ m((4π^2 lsinθ)/T^2 ) =(mg/cosθ)sinθ

F_t=mg/cosθ F_t=Mg
Mg=mg/cosθ
M=m/cosθ cosθ=m/M
∴ m((4π^2 lsinθ)/T^2 ) =(mg/(m⁄M ))sinθ
(m4π^2 lsinθ)/T^2 =Mgsinθ
(m4π^2 lsinθ)/Mgsinθ = T^2
(4π^2 lm)/Mg = T^2
T=√((4π^2 lm)/Mg)

Last edited: Oct 12, 2008
6. Oct 13, 2008

### tiny-tim

A bit long, but basically ok.