mass on rope attached to a rotating bar

1. is the linear momentum of the ball conserved in the $$e_3'$$ direction?

I figure since there is no force on this direction, but the direction is changing, the linear momentum will not be conserved.

1. is the angular momentum around B of the ball conserved in the $$e_3''$$ direction?

Here I can say that since $$\bar M=\bar x\times\bar F=-mgLsin(\theta)e_3''$$ the angular momentum in this direction is not conserved.

1. around the $$e_1''$$ direction?

Since in this direction the moment-force is 0 and it is fixed in space, the angular momentum is conserved. If $$\bar H=x_{A/B}\times \bar G$$ is the angular momentum, I find that $$\bar H \cdot e_1''=-m(Lsin(\theta))^2\dot \phi$$ conservation of $$\bar H$$ in this direction along with $$Lsin(\theta_0)\dot \phi_0=v_0$$ yields,$$\dot \phi=\frac{v_0sin(\theta)}{Lsin^2(\theta)}$$ note: In the formal answer to this question there is no square in the denominator, and I cannot see why.

1. Does the rope do work on the ball?

None, since it acts normal to it's direction.

1. Determine the kinetic energy of the ball.

Since no work is done by the rope and no other external forces, the following conservation of energy holds: $$E_0=E$$ $$E_0^{kinetic}+E_0^{potential}=E^{kinetic}+E^{potential}$$ $$mgL(1-cos(\theta))+\frac{1}{2}mv^2=const.$$ using this,$$\dot \phi$$ from before and $$v^2=(Lsin(\theta)\dot \phi)^2+(L\dot \theta)^2$$, $$\dot \theta$$ should be explicitly available.

1. Determine $$\dot\theta$$ and $$\dot\phi$$ in terms of {$$L,\theta,\theta_0,v_0,g$$}

$$\bar a=\frac{\delta \bar V}{\delta t}+\bar \omega\times\bar V$$ $$\bar V=\frac{\delta \bar X}{\delta t}+\bar \omega\times\bar X$$ where $$\frac{\delta}{\delta t}$$ is the frame-derivative and $$\bar \omega=e_1''(-\dot \phi)+e_3'(\dot \theta)$$ is the angular velocity of the ball's frame.
Using Newton's 2nd and 3rd law, and the acceleration above,$$T=-\bar F\cdot e_1'=-m\bar a\cdot e_1'$$