Consider the following setup
where the bead can glide along the rod without friction, and the rod rotates with a constant angular velocity $\omega$, and we want to find the constraint force using Lagrange multipliers. I chose the generalized coordinates $q=\{r,\varphi\}$ and the constraint equation $f$ to be $\varphi=\omega t$
We get the Lagrangian to be
$$\mathcal L= \frac 12m (\dot s^2 +s^2 \dot\varphi^2)- mgs\sin\varphi.$$
For the equation of motion, I got:
$$\begin{align} \frac {\partial\mathcal L}{\partial \varphi}&= -mgs\cos\varphi\\ \frac{\partial\mathcal L}{\partial\dot\varphi}&= ms^2\dot \varphi\\ \frac d{dt}\frac {\partial\mathcal L}{\partial\dot\varphi}&= 2ms\dot s\dot \varphi +ms^2\ddot\varphi\\ \Rightarrow\ 2ms\dot s \omega + mgs\cos(\omega t)&=\lambda\frac {\partial f}{\partial\varphi} = \lambda\\ \frac{\partial\mathcal L}{\partial s}&= ms\dot\varphi^2 -mg\sin\varphi\\ \frac d{dt}\frac{\partial\mathcal L}{\partial\dot s}&= m\ddot s\\ \Rightarrow\ m\ddot s &= ms\dot \varphi^2 =ms\dot\omega^2\ \Rightarrow\ s(t)=s_0 \cos(\omega t) \end{align}$$
and substituting $s(t)=s_0 \cos(\omega t)$ back to the equation for $\varphi$ results in:
$$-2ms_0^2\omega^2\cos(\omega t)\sin(\omega t) +mgs_0\cos^2(\omega t)=\lambda$$
and the constraint force is $C=\lambda\frac {\partial f}{\partial\varphi}+\lambda \frac{\partial f}{\partial s} = -2ms_0^2\omega^2\cos(\omega t)\sin(\omega t) +mgs_0\cos^2(\omega t)$
However, this is not true and the force is supposed to be $C= 2m\omega^2 s_0 \sin(\omega t)$, what am I doing wrong?
