2024-02-02 General Coordinates Continued

Uniform Circular motion

• You can use either Cartesian (x,y) or polar $(r,\theta )$coordinates
• Motion in x and y dependent on each other, 1 general coordinate
• Coordinate transform

$$x=r\cos \theta$$ $$y=r\sin \theta$$

forces of constraint

$$r^2=x^2+y^2$$

Particle confined to circle of radius $r$

• 2 Coordinates describe the position of the ball
• 1 force of constraint
• Therefore, 1 degree of freedom
• A suitable generalized coordinate is the polar coordinate $\theta$

Solving the Lagrangian

$$L=\frac{1}{2}m{v}^2$$ $$v={\dot{x}}^2+{\dot{y}}^2$$ $$\dot{x}=-r\dot{\theta }\sin \theta$$ $$\dot{y}=r\dot{\theta }\cos \theta$$ $$L=\frac{1}{2}m(r^2{\dot{\theta }}^2{\sin }^2\theta +r^2{\dot{\theta }}^2{\cos }^2\theta )$$ $$L=\frac{1}{2}m{r}^2{\dot{\theta }}^2$$ $$\frac{\partial L}{\partial \theta }=0$$ $$\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}=m{r}^2\ddot{\theta }$$ $$m{r}^2\ddot{\theta }=0\therefore \ddot{\theta }=0$$

Solving for init. conditions

$$\dot{\theta }=\text{ constant}=c_1$$ $$\theta =c_{1}t+c_2$$

Given:

$$\theta (t=0)={\theta }_0$$ $$\dot{\theta }(t=0)={\omega }_0$$ $$\theta (t)={\omega }_{0}t+{\theta }_0$$

Another example! Consider a ball of mass m on a uniformly rotating bar in a force-free space (no gravity, no friction). Bar is confined to rotate in the x-y plane

Physics: Ball will move radially outward due to centripetal acceleration.

Constraint: ball can only move on the bar $r(t)$

$$L=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)$$ $$x=r\cos (\omega t);\dot{x}=-r\omega \sin (\omega t)+\dot{r}\cos (\omega t)$$ $$y=r\sin (\omega t);\dot{y}=r\omega \cos (\omega t)+\dot{r}\sin (\omega t)$$ $$L=\frac{1}{2}m[r^2{\omega }^2{\sin }^2(\omega t)+\dot{r^2}{\cos }^2(\omega t)-2r\dot{r}\sin (\omega t)\cos (\omega t)+r^2{\omega }^2{\cos }^2(\omega t)\dots$$ $$+\dot{r^2}{\sin }^2(\omega t)+2r\dot{r}\omega \sin (\omega t)\cos (\omega t)$$ $$L=\frac{1}{2}m[{\dot{r}}^2+r^2\omega ]$$

Generalized Coordinate is r, so only 1 Euler-Lagrange Equation

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}=m\ddot{r}$$ $$\frac{\partial L}{\partial r}=mr{\omega }^2$$ $$m\ddot{r}-mr{\omega }^2=0$$ $$m\ddot{r}=mr{\omega }^2$$

This is force=centripetal acceleration

$$\omega =\frac{v}{r}\to {\omega }^2=\frac{v^2}{r^2}$$

Time to solve diffeq

$$r\propto e^{\omega t}$$ $$\dot{r}\approx \omega e^{\omega t}$$ $$\ddot{r}\approx {\omega }^2{e}^{\omega t}-{\omega }^{2}r$$ $$r(t)=A{e}^{\omega t}+B{e}^{-\omega t}$$ $$r(t=0)=r_0$$ $$\dot{r}(t=0)=0$$ $$\dot{r}(t)=A\omega e^{\omega t}-B\omega e^{-\omega t}$$ $$\ddot{r}(t)=A{\omega }^2{e}^{\omega t}+B{\omega }^2{e}^{-\omega t}$$ $${\omega }^{2}r(t)$$ $$r(t=0)=A{e}^0+B{e}^0=r_0$$ $$\dot{r}(t=0)=A\omega e^0-B\omega e^0=A\omega -B\omega =0\therefore A-B=0$$ $$A=B=\frac{r_0}{2}$$

Example: Rolling without slipping

Linear vs. rotational variables

$$v_t=R\dot{\theta }$$ $$ds=Rd\theta$$ $$v=\frac{ds}{dt}=R\frac{d\theta }{dt}=R\dot{\theta }$$ $$a_t=R_{\alpha }$$

Rolling without slipping implies that we can describe the system as the motion of the center of mass and the rotation about the center of mass. Rolling = translational + rotational

The object is doing 2 things.

Rolling without slipping $v_{com}=R\dot{\theta }=V_t$

Example: Consider a disk rolling down an inclined plane. Constraint of rolling without slipping, and $\varphi$ is constant. ($\varphi$ is the slope of the inclined plane).

Constraint is integrable!

$$\dot{x}=R\dot{\theta }=x=R\theta$$

Equation of constraint

$$x-R\theta =0$$

Kinetic energy:

$$T=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I{\dot{\theta }}^2$$

Potential energy:

$$u=-Mgx\sin (\varphi )$$ $$u=Mg(l-x)\sin \varphi$$ $$L=T-u=\frac{1}{2}M{\dot{x}}^2+\frac{1}{2}I{\dot{\theta }}^2-(g(l-x))\sin \varphi$$

1. We can substitute $\dot{\theta }=\frac{\dot{x}}{R}$ or $\dot{x}=R\dot{\theta }$
2. Use method of Lagrange Multipliers (cooler method) to change L to

$$\tilde{L}=\frac{1}{2}M{\dot{x}}^2+\frac{1}{2}I{\dot{\theta }}^2-Mg(l-x)\sin \theta +\lambda (x-R\theta )$$

We can go through this expression and come up with two equations of motion. There will be a $\lambda$ term in each of the equations of motion. We can use the constraint and $\lambda$ to solve and work through this problem. Use the constraints to destroy the constraints. $\lambda$ will be “something” that has meaning. yippee