# Homework Help: Mass revolving around a vertical axis

1. Feb 16, 2013

### Astrum

1. The problem statement, all variables and given/known data
Problem in the attachment

2. Relevant equations
-r(dθ/dt)^2= Centripetal force
F=ma

3. 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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Last edited: Feb 16, 2013
2. Feb 16, 2013

### haruspex

You left something out.
Let Tup, Tdown be the tensions. What equations can you write relating those to Fx and Fy?

3. Feb 16, 2013

### Astrum

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?

4. Feb 16, 2013

### haruspex

You defined Fx as a centripetal force. That means it is a resultant of other forces, so you should write -mrω2=Tx1+Tx2
That's ok, but now you have four unknowns. What equations derive Tx1 and Ty1 from T1?

5. Feb 16, 2013

### Astrum

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?

6. Feb 16, 2013

### haruspex

7. Feb 16, 2013

### Astrum

8. Feb 16, 2013

Yes.