Hint 1

(a) Each mass is moving with speed

$$v=\vert\hbox{{\boldmath $\omega$}}\times {\bf x}\vert=l\omega\sin\theta$$

around the axis of rotation. Each has an angular momentum of magnitude

$$L=m\vert{\bf x}\times{\bf v}\vert=ml^2\omega\sin\theta$$

pointing in a direction perpendicular to the rod in the same plane as ${\bf x}$ and $\hbox{{\boldmath$\omega$ }}$. Hence the total angular momentum is $L=2ml^2\omega\sin\theta$. It lies at an angle of ${\pi\over 2}-\theta$ to the axis of rotation.

(b) The torque needed to maintain this rotation has magnitude

$$\tau=\left\vert{d{\bf L}\over dt}\right\vert=\vert\omega\time... ...\sin ({\textstyle{\pi\over 2}}-\theta)=ml^2\omega^2\sin2\theta.$$

What is its direction?