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 the same meaning as the d in the figure.