How is angular momentum conserved in this proton+water scenario?

Consider a universe consisting of only one water molecule and one proton.

            H+            <----- p+
/
O-
\
H+


In the reference frame of the Oxygen, the proton is approaching one of the Hydrogen arms horizontally and the water molecule is not spinning. Total angular momentum of the system is zero.

The proton will repel the Hydrogen and vice-versa. The proton will decelerate and apply torque to the water molecule. It seems that total angular momentum is now nonzero.

What is wrong with this picture?

• There is nothing wrong besides your intuition about the changes in angular momentum. If you calculate the actual angular momentum then you will see that it's perfectly conserved. – CuriousOne Jun 15 '15 at 9:14
• If the proton is aimed at a hydrogen, then the initial angular momentum is not zero. – Javier Jun 15 '15 at 10:11
• Angular speed is initially zero, nothing is rotating. How can angular momentum be zero? – spraff Jun 15 '15 at 22:14
• Angular momentum doesn't have to involve something obviously "spinning". You can have orbital angular momentum. Basically, once you've chose an origin around which to compute angular momentum, any particle away from the origin that is not moving in the radial direction will have some contribution to angular momentum as per the formula $\vec L = \vec r \times \vec p$. – ApproximatelyTrue Jun 16 '15 at 8:18
• Take our origin to be the topmost hydrogen atom, so $r$ and $p$ are colinear, $L$ is zero. – spraff Jun 16 '15 at 8:32