Equation of motion in a disk and slider system I want to derive the equation of motion in this system: (the slider mass is m and the disk mass is M and the connecting bar is massless)

I have used relative velocity principle to calculate velocity of slider A:
$$\vec V_C=R\dot\theta \hat i $$
$$\vec V_B=\vec V_C+\vec V_{B/C} =R\dot\theta (1+\sin\theta) \hat i +R\dot\theta\cos\theta \hat j$$
$$\vec V_B=\vec V_A+\vec V_{B/A}=\vec V_A+2.5R\dot\phi\sin\phi \hat i -2.5R\dot\phi\cos\phi\hat j$$
Therfore:
$$\vec V_A=[R\dot\theta (1+\sin\theta)-2.5R\dot\phi\sin\phi] \hat i +[R\dot\theta\cos\theta+2.5R\dot\phi\cos\phi ]\hat j$$
And as we know the slider has no vertical motion so:
$$R\dot\theta\cos\theta+2.5R\dot\phi\cos\phi =0$$
$$\dot\theta\cos\theta=-2.5\dot\phi\cos\phi $$
Therefore:
$$\vec V_A=R\dot\theta (1+\sin\theta+\frac {\cos\theta}{\cos\phi})\hat i$$
From  geometry we know:
$$R\sin\theta =2.5R\sin\phi\Rightarrow \sin\theta =2.5\sin\phi$$
$$\cos\phi =\sqrt{1-\sin^2\phi}=\sqrt{1-\frac {1}{2.5^2}\sin^2\theta}=1+\frac{1}{25}\cos2\theta $$
If we want the acceleration in point A:
$$\vec a_A=\frac {d}{dt}\vec V_A=[R\ddot\theta (1+\sin\theta+\cos\theta)+R\dot\theta^2 (\cos\theta-\sin\theta)]\hat i$$
So the equation of motion can be derived using newton rule:
$$\sum \vec F=m\vec a $$
$$F (t)= mR\ddot\theta (1+\sin\theta+\cos\theta)+mR\dot\theta^2 (\cos\theta-\sin\theta)$$
Is my solution correct?
 A: It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
$$V=0$$
$$T=\frac{1}{2}mV_A^2+\frac{1}{2}I_{disk}\omega^2$$
$$I=\frac{3}{2}MR^2$$
if your answer for velocity of the slider is correct we can write:
$$T=\frac{1}{2}m[R\dot\theta (1+\sin\theta+\frac {\cos\theta}{\cos\phi})]^2+\frac{1}{2}\frac{3}{2}MR^2\dot\theta^2$$
so if you use Lagrange equations, you can find the answer:
$$L=T-V$$
$${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\partial L}{\partial {\dot {q}}_{j}}}\right)={\frac {\partial L}{\partial q_{j}}}}$$
A: The total kinetic energy is given by
$$
K = \frac 12 M||\vec v_C||^2+\frac 12 J_C \omega^2+\frac 12 m ||\vec v_A||^2
$$
with $\omega = \dot\theta$
We know also that
$$
\vec v_B = \vec v_C + \vec\omega\times(B-C)\\
\vec v_B = \vec v_A + \vec {\dot\phi} \times (B-A)
$$
or
$$
\vec v_C + \vec\omega\times(B-C) = \vec v_A + \vec {\dot\phi} \times (B-A)
$$
Here
$$
B-C = R(\cos\theta,\sin\theta)\\
B-A = \lambda R(-\cos\phi,\sin\phi)\\
\vec v_C = R(\omega,0)\\
\vec v_A = R (\omega-(\lambda\dot\phi+\omega)\sin\theta,(\omega-\lambda\dot\phi)\cos\theta)
$$
then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as
$$
T_{\eta}-\frac{d}{dt}T_{\dot\eta}=\mathcal{F}
$$
Here $\eta = (\theta,\phi)$
