Homework 3 Problem 9

#### Problem

Suppose that dissipation is introduced to the system in the previous problem with the Rayleigh dissipation function $R(r, \dot r, \theta, \dot \theta) = \frac{1}{2} b r \dot \theta^2$, where $b$is a positive constant. Compute the corresponding exterior force one form $\omega$ on $TQ$ and the vector field $X$ on $TQ$ satisfying the local Lagrange-d'Alembert principle $X\lrcorner \,\Omega _{L}-dE+\omega =0$.

#### Shu's solution.

Since the Rayleigh dissipation function is:

$R(r, \dot r, \theta, \dot \theta) = \frac{1}{2} b r \dot \theta^2$

The exterior force field is the fiber derivative of $-R$

$\mathbb{F}(-R):TQ{\large \ \rightarrow }T^{\ast }Q:(r,\theta ,\dot{r},\dot{\theta})\longmapsto (r,\theta ,\frac{\partial (-R)}{\partial \dot{r}},\frac{\partial (-R)}{\partial \dot{\theta}})=(r,\theta ,0,-br\dot{\theta})$

Then the exterior force one form $\omega$ satifies

$(\omega _{r}dr+\omega _{\theta }d\theta +\omega _{\dot{r}}d\dot{r}+\omega _{\dot{\theta}}d\dot{\theta})\cdot (v_{r}\frac{\partial }{\partial r}+v_{\theta }\frac{\partial }{\partial \theta }+v_{\dot{r}}\frac{\partial }{\partial \dot{r}}+v_{\dot{\theta}}\frac{\partial }{\partial \dot{\theta}})=\langle -br\dot{\theta}d\theta ,v_{r}\frac{\partial }{\partial r}+v_{\theta }\frac{\partial }{\partial \theta }\rangle$

Since $\omega$ is horizontal, then $\omega _{\dot{r}}=0$,$\omega _{\dot{\theta}}=0$

$\omega _{\theta }=-br\dot{\theta}$

$\omega =-br\dot{\theta}d\theta$

From problem 8, we get $E=\frac{1}{2}m\dot{r}^{2}+\frac{1}{2}mr^{2}\dot{\theta}^{2}+\frac{1}{2}kr^{2}$ and $\Omega _{L}=2mr\dot{\theta}dr\wedge d\theta -m\ddot{r}dr\wedge d\dot{r}-mr^{2}\ddot{\theta}d\theta \wedge d\dot{\theta}$

Then

$dE=(mr\dot{\theta}^{2}+kr)dr+m\dot{r}d\dot{r}+mr^{2}\dot{\theta}d\dot{\theta}$

$(dE-\omega )(w_{r}\frac{\partial }{\partial r}+w_{\theta }\frac{\partial }{\partial \theta }+w_{\dot{r}}\frac{\partial }{\partial \dot{r}}+w_{\dot{\theta}}\frac{\partial }{\partial \dot{\theta}})=(mr\dot{\theta}^{2}+kr)w_{r}+br\dot{\theta}w_{\theta }+m\dot{r}w_{\dot{r}}+mr^{2}\dot{\theta}w_{\dot{\theta}}$

$(X\lrcorner \Omega _{L})(w_{r}\frac{\partial }{\partial r}+w_{\theta }\frac{\partial }{\partial \theta }+w_{\dot{r}}\frac{\partial }{\partial \dot{r}}+w_{\dot{\theta}}\frac{\partial }{\partial \dot{\theta}})=\Omega _{L}(X_{r}\frac{\partial }{\partial r}+X_{\theta }\frac{\partial }{\partial \theta }+X_{\dot{r}}\frac{\partial }{\partial \dot{r}}+X_{\dot{\theta}}\frac{\partial }{\partial \dot{\theta}},w_{r}\frac{\partial }{\partial r}+w_{\theta }\frac{\partial }{\partial \theta }+w_{\dot{r}}\frac{\partial }{\partial \dot{r}}+w_{\dot{\theta}}\frac{\partial }{\partial \dot{\theta}})=(2mr\dot{\theta}dr\wedge d\theta -m\ddot{r}dr\wedge d\dot{r}-mr^{2}\ddot{\theta}d\theta \wedge d\dot{\theta})(X_{r}\frac{\partial }{\partial r}+X_{\theta }\frac{\partial }{\partial \theta }+X_{\dot{r}}\frac{\partial }{\partial \dot{r}}+X_{\dot{\theta}}\frac{\partial }{\partial \dot{\theta}},w_{r}\frac{\partial }{\partial r}+w_{\theta }\frac{\partial }{\partial�821�\theta }+w_{\dot{r}}\frac{\partial }{\partial \dot{r}}+w_{\dot{\theta}}\frac{\partial }{\partial \dot{\theta}})=(2mr\dot{\theta})(X_{r}w_{\theta }-w_{r}X_{\theta })-(m\ddot{r})(X_{r}w_{\dot{r}}-w_{r}X_{\dot{r}})-(mr^{2}\ddot{\theta})(X_{\theta }w_{\dot{\theta}}-w_{\theta }X_{\dot{\theta}})=(m\ddot{r}X_{\dot{r}}-2mr\dot{\theta}X_{\theta })w_{r}+(2mr\dot{\theta}X_{r}+mr^{2}\ddot{\theta}X_{\dot{\theta}})w_{\theta }-m\ddot{r}X_{r}w_{\dot{r}}-mr^{2}\ddot{\theta}X_{\theta }w_{\dot{\theta}}$

From the local Lagrange-d'Alembert principle $X\lrcorner \,\Omega _{L}-dE+\omega =0$, we get

$mr\dot{\theta}^{2}+kr=m\ddot{r}X_{\dot{r}}-2mr\dot{\theta}X_{\theta }$

$br\dot{\theta}=2mr\dot{\theta}X_{r}+mr^{2}\ddot{\theta}X_{\dot{\theta}}$

$m\dot{r}=-m\ddot{r}X_{r}$

$mr^{2}\dot{\theta}=-mr^{2}\ddot{\theta}X_{\theta }$

Then the the vector field $X=[X_{r},X_{\theta },X_{\dot{r}},X_{\dot{\theta}}]^{T}$ on $TQ$ is:

$X_{r}=-\frac{\dot{r}}{\ddot{r}}$

$X_{\theta }=-\frac{\dot{\theta}}{\ddot{\theta}}$

$X_{\dot{r}}=\frac{mr\dot{\theta}^{2}+kr-2mr\frac{\dot{\theta}^{2}}{\ddot{\theta}}}{m\ddot{r}}$

$X_{\dot{\theta}}=\frac{br\dot{\theta}+2mr\dot{\theta}\frac{\dot{r}}{\ddot{r}}}{mr^{2}\ddot{\theta}}$

page revision: 20, last edited: 15 May 2007 03:14