Mass revolving around a vertical axis

AI Thread Summary
The discussion focuses on solving a physics problem involving mass revolving around a vertical axis, specifically analyzing the forces acting on the mass. The participant breaks down the tension forces into components and sets up equations for centripetal force and gravitational force. They clarify the relationships between the tensions and the forces in both the x and y directions, ensuring to consider the angles of the tensions. The final expression for tension is derived, confirming the calculations are correct. The conversation emphasizes the importance of accurately defining forces and their directions in the problem-solving process.
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Homework Statement


Problem in the attachment


Homework Equations


-r(dθ/dt)^2= Centripetal force
F=ma


The Attempt at a Solution


The tension force will be cut into two parts.
I split the distance between the two strings into two sections, so the distance "r" to the mass can be defined as "L^2 - (d/2)^2 = r^2" where d is the total distance between the strings, and r is the distance to the mass.
F_{x}= -m\omega^{2}; F_{y} = -mg

The centripetal force will be divided by the two strings intuitively, but I'm not sure how to actually go about it mathematically. The two forces are already in their most basic component forms.

So, I have no idea how to deal with the vectors mathematically.

N.B. Sorry about the title, forgot to finish it =/
 

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Astrum said:
F_{x}= -m\omega^{2};
You left something out.
Let Tup, Tdown be the tensions. What equations can you write relating those to Fx and Fy?
 
haruspex said:
You left something out.
Let Tup, Tdown be the tensions. What equations can you write relating those to Fx and Fy?

I'm not sure what you mean. There are four forces involved. T1, T2, centripetal force and gravity.

The the two forces that are at a weird angle is the two tensions. The two tensions are both at a 45 degree angle from centripetal force.

Do you mean to rewrite it as: Fx= -mrω2+Tx1+Tx2=0
Fy= -mg + T1y+T2y=0

Split the two tensions into an x and y component?
 
Astrum said:
Fx= -mrω2+Tx1+Tx2=0
You defined Fx as a centripetal force. That means it is a resultant of other forces, so you should write -mrω2=Tx1+Tx2
Fy= -mg + T1y+T2y=0
That's ok, but now you have four unknowns. What equations derive Tx1 and Ty1 from T1?
 
T1x= T1cosθ, T1y=T1sinθ the same goes for T2

-mω2 = T1cosθ + T2cosθ
-mg + T2sinθ + T2sinθ = 0

Now I can solve the system of equations? Or am I missing something? :confused:
 
Astrum said:
T1x= T1cosθ, T1y=T1sinθ the same goes for T2

-mω2 = T1cosθ + T2cosθ
-mg + T2sinθ + T2sinθ = 0
Be careful with the signs. You are taking upward as positive. Which ways do the two tensions act in the vertical sense?
 
haruspex said:
Astrum said:
T1x= T1cosθ, T1y=T1sinθ the same goes for T2

-mω2 = T1cosθ + T2cosθ
-mg + T2sinθ + T2sinθ = 0
Be careful with the signs. You are taking upward as positive. Which ways do the two tensions act in the vertical sense?
-mg - T2sinθ + T1sinθ = 0

The second string will be downward. String 1 will have a positive y tension component, because it's balancing the y component from gravity.

So, solving the system, I get: T_{1}=\frac{mg}{sin(\theta)}-\frac{-m \omega^{2}}{cos(\theta)}

Does that look right?
 
Astrum said:
Does that look right?
Yes.
 
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