Mass revolving around a vertical axis

[Astrum]

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 =/

 

[haruspex]

$F_{x}= -m\omega^{2};$

You left something out.
Let T_up, T_down be the tensions. What equations can you write relating those to F_x and F_y?

 

[Astrum]

You left something out.
Let T_up, T_down be the tensions. What equations can you write relating those to F_x and F_y?

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: F_x= -mrω²+T_x1+T_x2=0
F_y= -mg + T_1y+T_2y=0

Split the two tensions into an x and y component?

 

[haruspex]

F_x= -mrω²+T_x1+T_x2=0

You defined F_x as a centripetal force. That means it is a resultant of other forces, so you should write -mrω²=T_x1+T_x2

F_y= -mg + T_1y+T_2y=0

That's ok, but now you have four unknowns. What equations derive T_x1 and T_y1 from T₁?

 

[Astrum]

T_1x= T₁cosθ, T_1y=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?

 

[haruspex]

T_1x= T₁cosθ, T_1y=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?

 

[Astrum]

T_1x= T₁cosθ, T_1y=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?

 

[haruspex]

Does that look right?

Yes.