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

In summary, the conversation is about deriving a formula for T, given the equations F_net = m((4π^2 r)/T^2 ) and F_net = F_c = 〖Ft〗_x = Ft×sinθ. The mistake preventing the formula from being solved was r = lsinθ, and after correcting it and combining the equations, the final formula for T is T=√((4π^2 lm)/Mg). The speaker also posted their calculations for verification.
  • #1
Epsillon
70
1
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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  • #2
Any help?
 
  • #3
Welcome to PF!

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

No, r = … ? :smile:
 
  • #4
actually r=lsin(pheta)
 
  • #5
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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  • #6
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:
 

1. How is the equation for circular motion derived?

The equation for circular motion can be derived using the concept of centripetal force and the Newton's second law of motion. The centripetal force is the force that keeps an object moving in a circular path and is always directed towards the center of the circle. By equating the centripetal force to the product of mass, velocity, and angular velocity, we can derive the equation for circular motion: Fc = mv²/r = mω²r.

2. What is the difference between uniform and non-uniform circular motion?

Uniform circular motion is when an object moves in a circular path at a constant speed, while non-uniform circular motion is when the speed of the object changes as it moves in a circular path. In non-uniform circular motion, the acceleration is not constant and changes both in magnitude and direction.

3. Is the centripetal force always directed towards the center of the circle?

Yes, the centripetal force is always directed towards the center of the circle. This is because the force is responsible for changing the direction of motion of an object, keeping it in a circular path. If the force is directed towards any other direction, the object will not be able to move in a circular path.

4. How is the centripetal force related to the radius of the circle?

The centripetal force is inversely proportional to the radius of the circle. This means that as the radius increases, the force decreases, and vice versa. This relationship can be seen in the equation for centripetal force, Fc = mv²/r. This also means that the centripetal force required to keep an object moving in a circular path at a certain speed decreases as the radius of the circle increases.

5. Can an object in circular motion have a constant speed and changing velocity?

Yes, an object in circular motion can have a constant speed and changing velocity. This is because velocity is a vector quantity that takes into account both magnitude and direction. In circular motion, the magnitude of the velocity remains constant, but the direction changes continuously, resulting in a changing velocity.

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