Angular momentum Conservation doubt Given an plane that has 2 pins attached to it. Each pin has a single arm firmly attached to it. Both of the pins can rotate on their axis. (The setup is similar to a clock).
We rotate the first pin in clockwise direction and the second pin in the anticlockwise direction at the same angular velocity starting at the same time. The arms on the pins would rotate with them. The plane itself is at rest.
Now, is it possible to start with 'any' two different angles that each arm makes w.r.t. the plane such that that despite the same velocity there is a motion in the plane?
My layman understanding says that due to conservation angular of momentum (since both arms have same velocity) there should be no movement in the plane. But, I was wondering if the initial angle of arms can make a difference (despite same magnitude of velocity).
 A: I'm going to answer the part I understand, which is whether the total angular momentum depends on any phase difference between the two hands.  I'll do this explicitly first, and then at the end say why it was a waste of time. For simplicity, I'll treat the hand as light, with a weight on the end of it.
Being in circular motion, the weight has a position
$$\vec{r} = A[\cos{(\omega t + \phi)}, \sin{(\omega t + \phi)}, 0],$$
where $\phi$ is some phase.  The velocity is then simply the time derivative of this expression,
$$\vec{p} = A m \omega[-\sin{(\omega t + \phi)}, \cos{(\omega t + \phi)}, 0].$$
Now the angular momentum is $\vec{L} = \vec{r}\times\vec{p}$, which upon computing the cross product gives
$$\vec{L} = A^2 m\omega [0, 0, \sin ^2(\omega t  +\phi )+ \cos ^2(\omega t +\phi )].$$
But we can recognise that final component is just equal to $1$, as an identity.  That is 
$$\vec{L} = A^2 m\omega[0,0,1].$$
And so the value of $\phi$ has made no difference to the value of $\vec{L}$.  Therefore each hand will contribute a certain angular momentum, and the phase difference makes on difference.
Why was this pointless?  We're working in a situation where we know angular momentum is conserved as the hands sweep around the clockface, hence we know to expect that angular momentum must be constant around the clock, and can not depend on $\phi$.  Since the total angular momentum is the sum of the angular momentum for each individual hand, we see that no, the phase difference makes on difference.
A: I assume that when you say "plane" you mean a "plate" which can slide on a frictionless horizontal surface. The pins are fixed to the plate, the arms are fixed at one end to the pins and rotate - either freely or driven by motors.
The resultant of all outside forces on the composite object is zero, so its centre of mass (CM) does not move. This is because of the conservation of linear momentum. There is no resultant torque on the object, so its total angular momentum is also conserved - even if the arms are driven by motors. This includes the angular momentum of the plate as well as that of the arms.
The CM of the plate will move if the CM of the two arms moves, so as to keep the overall CM in one place. Also, the plate could rotate. If the motors controlling the arms are switched on and the arms rotate in the same sense (clockwise or anticlockwise) then the plate will rotate in the opposite sense.
The rotation of the arms creates forces and possibly also a torque on the plate which cause it to move. If the arms move freely then the motion of the plate affect the motion of the arms. In general the interaction between the arms and the plate is complex, like that of a double pendulum - more so, because there are now three interacting parts. This motion could be chaotic. If the angular velocity of the arms relative to the plate is controlled using motors, the motion of the plate is predictable, not chaotic.
So generally the plate will move : it could translate and also rotate. Constraining forces (and torques) would be needed to keep it in place. If it is released it will move.

Assume that the rotation of the arms relative to the plate is controlled by motors, and is kept at the same constant angular velocity. Also that the plate is released from rest with the arms rotating.
Each rotating arm pulls outwards on its pin, in the direction of the arm. The direction of this force changes as the arm rotates. The resultant force on the plate is the vector sum of the forces from the two arms. It causes the centre of mass (CM) of the plate to move. If the arms point in directions which are more than $90^{\circ}$ apart, then there is also a torque on the plate, causing it to rotate about its CM. 
If the two arms rotate in the same sense (clockwise or anticlockwise) then the angle between their directions (the phase difference $\phi$) is fixed. The resultant force on the plate is constant in magnitude but its direction rotates. The magnitude of this force is maximum when $\phi=0^{\circ}$ and zero when $\phi=180^{\circ}$. The CM of the plate moves in a circle. If there is a torque (ie $\phi \ge 90^{\circ}$ then it changes magnitude and sense (clockwise/anticlockwise) as the arms rotate. Because of this the plate oscillates, turning clockwise then anticlockwise. 
If the arms rotate in opposite senses then the angle between them $\phi$ must change. The resultant force on the plate therefore changes in magnitude as well as direction. However, the direction is always in a straight line, so the CM of the plate oscillates along a straight line. This line will be the direction in which they are parallel or anti-parallel. 
