I believe I have solved below problem. I am not looking for help on problem-solving per se. I am just looking for an intuitive explanation.
Problem statement: wheel mass = $m_1$, even mass rod BC mass = $m_2$. rod length = $2l$. Rod hingers on wheel so wheel can move freely. rod started falling due to a small perturbation. Wheel started rolling without slipping (so there is a frictional force, but the work of the frictional force can be ignored). When rod is at horizontal position, what's the velocity of wheel C and the angular velocity of the stick w.r.t. C?
Let $x$ be the coordinate of the wheel. $\theta$ be the angle between the rod and the vertical line. Let $A$ be the center of $BC$. Then
$x_A = x + l \sin \theta$, $y_A = l \cos \theta$, $\dot{x_A} = \dot{x} + l \cos \theta \dot{\theta}$, $\dot{y_A} = -l \sin \theta \dot{\theta}$
Kinetic energy consists of the horizontal movement of the wheel, the falling rod, and the rotation of the wheel:
\begin{align} T & = \frac{1}{2} m_v v_C^2 + \frac{1}{2} \frac{1}{2} m_1 R^2 (\frac{v_C}{R})^2 + \frac{1}{2} m_v v_A^2 + \frac{1}{2} \frac{1}{12} m_2 (2l)^2 \dot{\theta}^2\\ & = (\frac{3}{4} m_1 + \frac{1}{2} m_2) \dot{x}^2 + \frac{2}{3} m_2 l^2 \dot{\theta}^2 + m_2 l \dot{\theta} \dot{x} \cos \theta\\ \end{align}
$$V = m_2 g \cos \theta l$$
$$T+V = m_2 gl$$
It looks intimidating, but here is a trick. $L = T-V$ doesn't have $x$ so from Lagrange Equation, $$\frac{d}{dt} (\frac{\partial L}{\partial \dot {x}}) = \frac{\partial L}{\partial x} = 0$$
we have $\frac{\partial L}{\partial \dot{x}} = C $ a constant. So we can get $\dot{x} = 0, \dot{\theta} = \sqrt{\frac{3g}{2l}}$, when $\theta = 90^{\circ}$.
But is there an intuitive explanation of why $\dot{x} = 0$?
If this is not a wheel, but rather a block or a cart on a frictionless surface, then this explanation would be trivial. The momentum conserves on the horizontal direction.
But since there is a wheel that's rolling without slipping, there is always a frictional force to one direction from $\theta = 0$ to $\theta = 90$, and we can't use the conservation of the momentum here. How do we intuitively explain $\dot{x} = 0$?
