Ball Bearing Inside a Hollow, Spinning Rod: where is the logical flaw? As described in the title, suppose we have a frictionless, hollow rod that is rotating in the $xy$-plane with some fixed angular velocity $\omega$. The rod is pivoting around its midpoint. Suppose we place a ball bearing inside of this rod such that the diameter of the ball bearing is equal to the diameter of the hollow opening of the rod. Additionally, assume plane polar coordinates: $(r,\phi)$. 
So, the constraint equation is simply $$G(\phi,t) = \phi - \omega t = 0.$$ Now, suppose that we want to find the $\phi$ component of the constraint force on the ball bearing. I know that blindly applying the generalized Lagrange equations yields 
$$ \lambda = 2mr\dot{r}\dot{\phi},$$
which then implies
$$ F_{\phi,constraint} = \lambda \frac{\partial G}{\partial \phi} = 2mr\dot{r}\dot{\phi},$$
but I am not certain that time-dependent constraints can be treated in the same way. What exactly is incorrect with the following argument?

The bead is constrained to rotate in the $xy$-plane with constant
  angular velocity $\omega$, hence $\ddot{\phi} = 0$. Thus, since there
  is no angular acceleration we know that the net force on the bead in
  the $\phi$ direction must be zero. However, the only force on the bead
  in the $\phi$ direction is the constraint force from the rod on the
  bead. Therefore, the $\phi$ component of the constraint force is zero.

I want to justify this by saying that the bead feels equal and opposite $\phi$ forces from the rod, as it is surrounded on all sides and thus the net force would be zero, but I think there is something fundamentally wrong with this argument. Any insight is greatly appreciated. 
 A: You seem to realize that the rotational analog for Newton's law is important here. This law states that the net torque $\tau$ on an object and its angular momentum $L$ are related by $\tau = \dot{L}$. 
If I read you question correctly you seem to think that because $\ddot{\phi}=0$, that the angular momentum $L$ must be constant. However, this isn't true. Recall that the angular momentum $L$ is given by $L=mr^2\dot{\phi}$, where $m$ is the mass of the bead and $r$ is its radius. Now in this case the radius of the bead $r$ is changing while $\dot{\phi}$ and $m$ stay constant. Therefore the angular momentum $L$ is in fact changing. So the flaw in the argument is saying that $\ddot{\phi}=0$ implies $\dot{L}=0$.
A: The generalized coordinates are the 2 polar coordinates $r$ and $\phi$. There is 1 constraint. This means we have 2 Lagrange equations of first kind with a constraint force term:


*

*Lagrange equation for $r$ is the Newton's 2nd law in the radial direction with a centrifugal force and no constraint force.

*Lagrange equation for $\phi$ is the angular Newton's 2nd law $\dot{L}=\lambda$, where $\lambda$ is the generalized force of the constraint, in this case a torque. Here $L=mr^2\omega$ is the angular momentum.
As OP observes, there is no angular acceleration. This because the constraint force $\frac{\lambda}{r}$ is balanced by the Coriolis force $2m\dot{r}\omega$ in the rotating reference frame.
