# Movement of inclined plane with its slope changing over time

Imagine a situation of a mass over the center of inclined planed which its slope in changing over time. The angle changes acording to the movement of forced damped harmonic oscillator. I want to know the position of the particle respect the center, so i use polar coordinates $$x=r\cos\phi$$ and $$y=r\sin\phi$$ when $$\phi$$ is the solution of forced damped harmonic oscillator $$a\ddot\phi+b\dot\phi+c\phi=F\sin\omega t$$ that gives me the angle of the slope.

The lagrangian is given by:

$$L=\frac{1}{2}m(\dot r^2+r^2\dot\phi^2)-mgr\sin\phi$$

On the other hand, I want to consider the dissipative force between the mass and the plane so the Rayleigh dissipation function is $$R=-\mu Nv=-\mu mg\cos\phi \sqrt{\dot r^2+r^2\dot\phi^2}$$

Then, the equation of motion is given by:

$$\frac{d}{dt}\frac{\partial L}{\partial \dot r}-\frac{\partial L}{\partial r}+\frac{\partial R}{\partial \dot r}=0$$

The angle $$\phi$$ is a constraint given by the damped equation and i know its value every time.

Finally the motion is given by:

$$\ddot r-\mu g\cos\phi\frac{\dot r}{\sqrt{\dot r^2+r^2\dot\phi^2}}-\dot\phi^2r+g\sin\phi=0$$

My question is if the approach I have done of the problem is correct.

$$\mathbf R=\left[ \begin {array}{c} s\sin \left( \phi \left( \tau \right) \right) \\ H-s\cos \left( \phi \left( \tau \right) \right) \end {array} \right]$$

from here the velocity is:

$$\mathbf v=\left[ \begin {array}{c} \sin \left( \phi \left( \tau \right) \right) {\dot s}+s\cos \left( \phi \left( \tau \right) \right) { \frac {d}{d\tau}}\phi \left( \tau \right) \\ -\cos \left( \phi \left( \tau \right) \right) {\dot s}+s\sin \left( \phi \left( \tau \right) \right) {\frac {d}{d\tau}}\phi \left( \tau \right) \end {array} \right]$$

hence the kinetic energy

$$T=\frac m2\,\mathbf{v}\cdot\mathbf{v}$$

the potential energy

$$U=m\,g\,\mathbf R_y-F_\mu\,s$$ where $$~F_\mu=\mu\,N~$$ is the friction force

and with EL you obtain the EOM

$$m\,\frac{d^2\,s}{d\tau^2}-m \left( {\frac {d}{d\tau}}\phi \left( \tau \right) \right) ^{2}\,s(\tau)-m\,g\cos \left( \phi \left( \tau \right) \right) +\mu\,N =0\\ N=m\,g\sin(\phi(\tau))$$

the function $$~\phi(\tau)~$$ is the solution of damped harmonic oscillator.

• Thanks a lot!! I was annoying with my firctional term in the equation because it was so complicated, i reaize that the dissipative force only consider the S velocity of Rayleigh function a not de velocity of Y too. Oct 11, 2021 at 14:11
• you can use this friction equation $~-\mu\,|N|\,\frac{v_t}{|v_t|}$ where $~v_t~$ is the velocity towards the friction force
– Eli
Oct 11, 2021 at 15:03
• I see , your variable r is my variable s, the EOM are equal !!
– Eli
Oct 11, 2021 at 15:04
• Yes, i did not realize that i was using all components of velocity instead of tangential velocity. Thanks for the feedback! Oct 13, 2021 at 8:23
• I realize that if $g cos\phi<\dot\phi s$ and $cos\phi<0$, the movement would go further and further, so the friction term $\mu N$ should it be $-\mu N$? Oct 14, 2021 at 14:15