How do I correctly derive the formula for circular motion in a physics lab?

AI Thread Summary
The discussion focuses on deriving the formula for circular motion, specifically T=√((4π^2 lm)/Mg). The user initially misidentified the radius as r=lcosθ, but corrected it to r=lsinθ, which was crucial for solving the problem. They derived the net force equations and combined them to reach the final formula. The user shared their derivation process for verification and expressed willingness to provide additional clarification if needed. The overall conclusion is that the derivation is correct, confirming the user's understanding of the topic.
Epsillon
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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
 
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Any help?
 
Welcome to PF!

Hi Epsillon! Welcome to PF! :smile:
Epsillon said:
… and I know that r=lcos …

No, r = … ? :smile:
 
actually r=lsin(pheta)
 
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)
 
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Epsillon said:
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?

A bit long, but basically ok. :smile:
 
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