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

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Homework Help Overview

The discussion revolves around deriving the formula for circular motion, specifically T=√((4π^2 lm)/Mg), in a physics lab context. Participants are exploring the relationships between forces acting on an object in circular motion and the parameters involved in the derivation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the net force acting on the object and its relation to centripetal acceleration. There is an initial confusion regarding the correct expression for the radius in terms of the length and angle, leading to a reevaluation of the equations used.

Discussion Status

Some participants have made progress in their derivation and shared their findings for verification. There is an acknowledgment of a previous mistake regarding the expression for radius, which has been corrected. However, the discussion remains open for further validation and clarification of the steps taken.

Contextual Notes

Participants are working under the constraints of a lab assignment, which may impose specific requirements or assumptions about the setup and parameters involved in the derivation.

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
 
Last edited:
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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)
 
Last edited:
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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