# Angular momentum conservation

**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 T_{1} be the tension when θ_{1}=30º, m*g be the weight and v_{1} be the velocity. Then we can get tg30º = (m*(v_{1})²/R_{1})/(mg) = √3/3 => √3/3*g=v²_{1}/R_{1} (I)

and when Θ=60º we have √3g=v²_{2}/R_{2}. (II)

From these equations I divided I by II:

1/3= (v_{1}/v_{2})²*R_{2}/R_{1} and obtained (III). But we threen unknows (v_{2}, R_{2}, v_{1}) 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 the same meaning as the d in the figure.

http://ift.tt/1e3fgEF

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