Equation of motion in a disk and slider system

Question

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?

Answer 1

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: $$L=T-V$$

$${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\partial L}{\partial {\dot {q}}_{j}}}\right)={\frac {\partial L}{\partial q_{j}}}}$$

Answer 2

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 $\eta = (\theta,\phi)$