# Angular momentum conservation

1. Feb 22, 2014

### Icaro Amorim

1. The problem statement, all variables and given/known data
Hello, guys. I'm not a native english speaker, so don't mind if I wrote something incorrectly (if you find any error or difficulty in what I wrote just inform me, please).

A little ball is hold by a thread of negligible mass which moves round a vertical axis with constant velocity. It keeps a distance of 0,5m from the axis when the anglo θ is equal 30º (as shown in the figure uploaded). The thread goes through a little hole O in a slab and it is slowly pulled up until the angle Θ becomes 60º. What is the thread's lenght pulled? What is the variation factor of velocity?

2. Relevant equations
L = rxmv, T*senΘ=mv²/R and T*cosΘ-mg = 0

3. The attempt at a solution
I've tried solving this question this way:
"let T1 be the tension when θ1=30º, m*g be the weight and v1 be the velocity. Then we can get tg30º = (m*(v1)²/R1)/(mg) = √3/3 => √3/3*g=v²1/R1 (I)
and when Θ=60º we have √3g=v²2/R2. (II)
From these equations I divided I by II:
1/3= (v1/v2)²*R2/R1 and obtained (III). But we threen unknows (v2, R2, v1) and only one equation.

I imagine the angular momentum might conserve in the perpendicular direction to the plane of motion of the mass once it is slowly pulled up, but I have no idea how to apply this.

P.S.: I used R but it is the same in meaning as d in the figure.

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Last edited: Feb 22, 2014
2. Feb 22, 2014

### voko

You have two equations, (I) and (II). Add to them conservation of angular momentum, and you will have three equations and three unknowns. Solve.

3. Feb 22, 2014

### Icaro Amorim

That's the problem. I don't know how to apply conservation of angular momentum. I tried M*v1*(R1/tgθ1) = M*v2*(R2/tgθ2) where R/tgθ is the distance from the point O to the plane of motion and it was incorrect. Can you tell me why?

4. Feb 22, 2014

### voko

You said that $R$ was the same as $d$ in the figure. Then angular momentum is simply $m v_1 R_1$ and $m v_2 R_2$.