### 3.2Second HW set, Solve the ﬁrst set using Lagrangian

3.2.1  problem 1
3.2.2  Decouple the ODE’s
3.2.3  problem 2
3.2.4  Solution problem 3
3.2.5  problem 4
3.2.6  Solution problem 5
3.2.7  Key solution

solve the ﬁrst practice problems using Lagrangian approach.

#### 3.2.1problem 1

##### Velocity diagram

Only the velocity diagram is needed for the Lagrangian method. The generalized coordinates are: $$x,\theta$$

Let $$L$$ be the Lagrangian, and let $$T$$ be the kinetic energy of the system, and $$V$$ the potential energy.$L=T-V$$T=\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}mv^{2}$ Where $$v$$ is the speed of the mass $$m$$ which is

Hence\begin{align*} T & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \left ( \dot{x}+L\dot{\theta }\cos \theta \right ) ^{2}+\left ( L\dot{\theta }\sin \theta \right ) ^{2}\right ) \\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}^{2}+\left ( L\dot{\theta }\right ) ^{2}+2\dot{x}L\dot{\theta }\cos \theta \right ) \end{align*}

For the potential energy$V=\frac{1}{2}kx^{2}+mgL\cos \theta$ Therefore, $$L$$ becomes\begin{align*} L & =T-V\\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}^{2}+\left ( L\dot{\theta }\right ) ^{2}+2\dot{x}L\dot{\theta }\cos \theta \right ) -\left ( \frac{1}{2}kx^{2}+mgL\cos \theta \right ) \end{align*}

To obtain equations of motion, we evaluate $\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{x}}\right ) -\frac{\partial L}{\partial x}=Q_{x}$

and$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{\theta }}\right ) -\frac{\partial L}{\partial \theta }=Q_{\theta }$

For the generalized forces, we can readily see that $$Q_{x}=F$$ and $$Q_{\theta }=0$$. Hence for $$x$$ we write

$\frac{\partial L}{\partial \dot{x}}=M\dot{x}+m\dot{x}+mL\dot{\theta }\cos \theta$

$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{x}}\right ) =M\ddot{x}+m\ddot{x}+mL\ddot{\theta }\cos \theta -mL\dot{\theta }^{2}\sin \theta$

$\frac{\partial L}{\partial x}=-kx$ Hence, for the mass $$M$$, the equation of motion is

To ﬁnd EQM for mass $$m$$, evaluate $$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{\theta }}\right ) -\frac{\partial L}{\partial \theta }=0$$$\frac{\partial L}{\partial \dot{\theta }}=mL^{2}\dot{\theta }+m\dot{x}L\cos \theta$ $\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{\theta }}\right ) =mL^{2}\ddot{\theta }+m\ddot{x}L\cos \theta -m\dot{x}L\dot{\theta }\sin \theta$ $\frac{\partial L}{\partial \theta }=-m\dot{x}L\dot{\theta }\sin \theta +mgL\sin \theta$ Hence EQM is\begin{align} \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{\theta }}\right ) -\frac{\partial L}{\partial \theta } & =0\nonumber \\ mL^{2}\ddot{\theta }+m\ddot{x}L\cos \theta -m\dot{x}L\dot{\theta }\sin \theta -\left ( -m\dot{x}L\dot{\theta }\sin \theta +mgL\sin \theta \right ) & =0\nonumber \\ L\ddot{\theta }+\ddot{x}\cos \theta -\dot{x}\dot{\theta }\sin \theta +\dot{x}\dot{\theta }\sin \theta -g\sin \theta & =0\nonumber \\ L\ddot{\theta }+\ddot{x}\cos \theta -g\sin \theta & =0\nonumber \\ \ddot{\theta } & =\frac{g\sin \theta -\ddot{x}\cos \theta }{L}\tag{2} \end{align}

#### 3.2.2Decouple the ODE’s

Substitute Eq. (2) into Eq. (1) gives\begin{align} x^{\prime \prime } & =\frac{F\left ( t\right ) -kx-mL\left [ \frac{g\sin \theta -x^{\prime \prime }\cos \theta }{L}\right ] \cos \theta +mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{M+m}\nonumber \\ & =\frac{F\left ( t\right ) -kx-mg\sin \theta \cos \theta +mx^{\prime \prime }\cos ^{2}\theta +mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{M+m}\nonumber \\ x^{\prime \prime }\left ( M+m\right ) -mx^{\prime \prime }\cos ^{2}\theta & =F\left ( t\right ) -kx-mg\sin \theta \cos \theta +mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \nonumber \\ x^{\prime \prime } & =\frac{F\left ( t\right ) -kx-mg\sin \theta \cos \theta +mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{\left ( M+m-m\cos ^{2}\theta \right ) }\tag{3} \end{align}

Also, we can solve for $$\theta ^{\prime \prime }$$ by Substituting Eq. (1) into (2)\begin{align} \theta ^{\prime \prime } & =\frac{g\sin \theta -\left ( \frac{F\left ( t\right ) -kx-mL\theta ^{\prime \prime }\cos \theta +mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{M+m}\right ) \cos \theta }{L}\nonumber \\ L\theta ^{\prime \prime }\left ( M+m\right ) & =g\sin \theta \left ( M+m\right ) -\left ( F\left ( t\right ) -kx-mL\theta ^{\prime \prime }\cos \theta +mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \right ) \cos \theta \nonumber \\ L\theta ^{\prime \prime }\left ( M+m\right ) & =g\sin \theta \left ( M+m\right ) -F\left ( t\right ) \cos \theta +kx\cos \theta +mL\theta ^{\prime \prime }\cos ^{2}\theta -mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \cos \theta \nonumber \\ L\theta ^{\prime \prime }\left ( M+m\right ) -mL\theta ^{\prime \prime }\cos ^{2}\theta & =g\sin \theta \left ( M+m\right ) -F\left ( t\right ) \cos \theta +kx\cos \theta -mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \cos \theta \nonumber \\ \theta ^{\prime \prime } & =\frac{g\sin \theta \left ( M+m\right ) -F\left ( t\right ) \cos \theta +kx\cos \theta -mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \cos \theta }{L\left ( M+m-m\cos ^{2}\theta \right ) }\tag{4} \end{align}

Eqs. (3) and (4) is what we will use to convert the system to ﬁrst order form since they are in decoupled form.

##### Convert to state space form

Using the decoupled ODE’s above, Eqs (3) and (4), and introducing 4 state variables $$x_{1},x_{2},x_{3}\,x_{4}$$ we obtain\begin{align*} \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} & =\begin{pmatrix} x\\ \theta \\ x^{\prime }\\ \theta ^{\prime }\end{pmatrix} \overset{\frac{d}{dt}}{\rightarrow }\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} =\begin{pmatrix} x^{\prime }\\ \theta ^{\prime }\\ x^{\prime \prime }\\ \theta ^{\prime \prime }\end{pmatrix} \\\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} & =\begin{pmatrix} x_{3}\\ x_{4}\\ \frac{F\left ( t\right ) -kx-mg\sin \theta \cos \theta +mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{\left ( M+m-m\cos ^{2}\theta \right ) }\\ \frac{g\sin \theta \left ( M+m\right ) -F\left ( t\right ) \cos \theta +kx\cos \theta -mL\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \cos \theta }{L\left ( M+m-m\cos ^{2}\theta \right ) }\end{pmatrix} \\\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} & =\begin{pmatrix} x_{3}\\ x_{4}\\ \frac{F\left ( t\right ) -kx_{1}-mg\sin x_{2}\cos x_{2}+mLx_{4}^{2}\sin x_{1}}{\left ( M+m-m\cos ^{2}x_{2}\right ) }\\ \frac{g\sin x_{2}\left ( M+m\right ) -F\left ( t\right ) \cos x_{2}+kx_{1}\cos x_{2}-mLx_{4}^{2}\sin x_{2}\cos x_{2}}{L\left ( M+m-m\cos ^{2}x_{2}\right ) }\end{pmatrix} \end{align*}

#### 3.2.3problem 2

##### Velocity diagram

Only the velocity diagram is needed for the Lagrangian method. The generalized coordinates are: $$x,\theta$$

Let $$L$$ be the Lagrangian, and let $$T$$ be the kinetic energy of the system, and $$V$$ the potential energy.$L=T-V$$T=\frac{1}{2}M\dot{x}^{2}+\overset{\text{angular bar K.E.}}{\overbrace{\frac{1}{2}I_{g}\dot{\theta }^{2}}}+\overset{\text{linear\ bar\ K.E. of its c.g.}}{\overbrace{\frac{1}{2}m\left ( \dot{x}^{2}+\left ( a\dot{\theta }\right ) ^{2}+2\dot{x}a\dot{\theta }\cos \theta \right ) }}$ And$V=\frac{1}{2}kx^{2}+mga\cos \theta$ Therefore, $$L$$ becomes\begin{align*} L & =T-V\\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}I_{g}\dot{\theta }^{2}+\frac{1}{2}m\left ( \dot{x}^{2}+\left ( a\dot{\theta }\right ) ^{2}+2\dot{x}a\dot{\theta }\cos \theta \right ) -\left ( \frac{1}{2}kx^{2}+mga\cos \theta \right ) \end{align*}

To obtain equations of motion, we evaluate $\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{x}}\right ) -\frac{\partial L}{\partial x}=F$ and$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{\theta }}\right ) -\frac{\partial L}{\partial \theta }=0$ For $$M$$ we obtain$\frac{\partial L}{\partial \dot{x}}=M\dot{x}+m\dot{x}+ma\dot{\theta }\cos \theta$$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{x}}\right ) =M\ddot{x}+m\ddot{x}+ma\ddot{\theta }\cos \theta -ma\dot{\theta }^{2}\sin \theta$$\frac{\partial L}{\partial x}=-kx$ Hence, for the mass $$M$$, the equation of motion is\begin{align} M\ddot{x}+m\ddot{x}+ma\ddot{\theta }\cos \theta -ma\dot{\theta }^{2}\sin \theta +kx & =F\nonumber \\ \ddot{x} & =\frac{F-kx-ma\ddot{\theta }\cos \theta +ma\dot{\theta }^{2}\sin \theta }{M+m}\tag{1} \end{align}

To ﬁnd EQM for mass $$m$$, evaluate $$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{\theta }}\right ) -\frac{\partial L}{\partial \theta }=0$$$\frac{\partial L}{\partial \dot{\theta }}=I_{g}\dot{\theta }+m\left ( a^{2}\dot{\theta }+\dot{x}a\cos \theta \right )$

$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{\theta }}\right ) =I_{g}\ddot{\theta }+m\left ( a^{2}\ddot{\theta }-\dot{x}a\dot{\theta }\sin \theta +\ddot{x}a\cos \theta \right )$

$\frac{\partial L}{\partial \theta }=-m\dot{x}a\dot{\theta }\sin \theta +mga\sin \theta$ Hence EQM is\begin{align} \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{\theta }}\right ) -\frac{\partial L}{\partial \theta } & =0\nonumber \\ I_{g}\ddot{\theta }+m\left ( a^{2}\ddot{\theta }-\dot{x}a\dot{\theta }\sin \theta +\ddot{x}a\cos \theta \right ) +m\dot{x}a\dot{\theta }\sin \theta -mga\sin \theta & =0\nonumber \\ I_{g}\ddot{\theta }+ma^{2}\ddot{\theta }-m\dot{x}a\dot{\theta }\sin \theta +m\ddot{x}a\cos \theta +m\dot{x}a\dot{\theta }\sin \theta -mga\sin \theta & =0\nonumber \\ I_{g}\ddot{\theta }+ma^{2}\ddot{\theta }+m\ddot{x}a\cos \theta -mga\sin \theta & =0\nonumber \\ \ddot{\theta }\left ( I_{g}+ma^{2}\right ) & =mga\sin \theta -m\ddot{x}a\cos \theta \nonumber \\ \ddot{\theta } & =\frac{mga\sin \theta -m\ddot{x}a\cos \theta }{I_{g}+ma^{2}}\nonumber \\ & =\frac{mag\sin \theta -max^{\prime \prime }\cos \theta }{ma^{2}+I_{g}}\tag{2} \end{align}

##### Decouple the ODE’s

Substitute Eq. (2) into Eq. (1) gives$x^{\prime \prime }=\frac{F\left ( t\right ) -kx-ma\left [ \frac{mag\sin \theta -max^{\prime \prime }\cos \theta }{ma^{2}+I_{g}}\right ] \cos \theta +ma\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{M+m}$ Let $$\left ( ma^{2}+I_{g}\right ) =I_{o}$$, hence\begin{align} x^{\prime \prime } & =\frac{I_{o}F\left ( t\right ) -I_{o}kx-ma\left [ mag\sin \theta -max^{\prime \prime }\cos \theta \right ] \cos \theta +I_{o}ma\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{I_{o}\left ( M+m\right ) }\nonumber \\ I_{o}\left ( M+m\right ) x^{\prime \prime }-m^{2}a^{2}x^{\prime \prime }\cos \theta & =I_{o}F\left ( t\right ) -I_{o}kx-m^{2}a^{2}g\sin \theta \cos \theta +I_{o}ma\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \nonumber \\ x^{\prime \prime } & =\frac{I_{o}F\left ( t\right ) -I_{o}kx-m^{2}a^{2}g\sin \theta \cos \theta +I_{o}ma\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{I_{o}\left ( M+m\right ) -m^{2}a^{2}\cos \theta }\tag{3} \end{align}

Also, we can solve for $$\theta ^{\prime \prime }$$ by Substituting Eq. (1) into (2)$\theta ^{\prime \prime }=\frac{mag\sin \theta -ma\left [ \frac{F\left ( t\right ) -kx-ma\theta ^{\prime \prime }\cos \theta +ma\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{M+m}\right ] \cos \theta }{ma^{2}+I_{g}}$ Let $$\left ( ma^{2}+I_{g}\right ) =I_{o}$$, hence

Therefore, the ﬁnal EQM are\begin{align*} x^{\prime \prime } & =\frac{I_{o}F\left ( t\right ) -I_{o}kx-m^{2}a^{2}g\sin \theta \cos \theta +I_{o}ma\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{I_{o}\left ( M+m\right ) -m^{2}a^{2}\cos \theta }\\ \theta ^{\prime \prime } & =\frac{\left ( M+m\right ) mag\sin \theta -maF\left ( t\right ) \cos \theta +makx\cos \theta -m^{2}a^{2}\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \cos \theta }{I_{o}\left ( M+m\right ) -m^{2}a^{2}\cos ^{2}\theta } \end{align*}

##### Convert to state space form

Using the decoupled ODE’s above, Eqs (3) and (4), and introducing 4 state variables $$x_{1},x_{2},x_{3}\,x_{4}$$ we obtain\begin{align*} \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} & =\begin{pmatrix} x\\ \theta \\ x^{\prime }\\ \theta ^{\prime }\end{pmatrix} \overset{\frac{d}{dt}}{\rightarrow }\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} =\begin{pmatrix} x^{\prime }\\ \theta ^{\prime }\\ x^{\prime \prime }\\ \theta ^{\prime \prime }\end{pmatrix} \\\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} & =\begin{pmatrix} x_{3}\\ x_{4}\\ \frac{I_{o}F\left ( t\right ) -I_{o}kx-m^{2}a^{2}g\sin \theta \cos \theta +I_{o}ma\left ( \theta ^{\prime }\right ) ^{2}\sin \theta }{I_{o}\left ( M+m\right ) -m^{2}a^{2}\cos \theta }\\ \frac{\left ( M+m\right ) mag\sin \theta -maF\left ( t\right ) \cos \theta +makx\cos \theta -m^{2}a^{2}\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \cos \theta }{I_{o}\left ( M+m\right ) -m^{2}a^{2}\cos ^{2}\theta }\end{pmatrix} \\\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} & =\begin{pmatrix} x_{3}\\ x_{4}\\ \frac{I_{o}F\left ( t\right ) -I_{o}kx_{1}-m^{2}a^{2}g\sin x_{2}\cos x_{2}+I_{o}ma\ x_{4}^{2}\sin x_{2}}{I_{o}\left ( M+m\right ) -m^{2}a^{2}\cos x_{2}}\\ \frac{\left ( M+m\right ) mag\sin x_{2}-maF\left ( t\right ) \cos x_{2}+makx\cos x_{2}-m^{2}a^{2}x_{4}^{2}\sin x_{2}\cos x_{2}}{I_{o}\left ( M+m\right ) -m^{2}a^{2}\cos ^{2}x_{2}}\end{pmatrix} \end{align*}

#### 3.2.4Solution problem 3

##### Velocity diagram

Only the velocity diagram is needed for the Lagrangian method. Generalized coordinates are: $$x,\theta ,r$$

Let $$L$$ be the Lagrangian, and let $$T$$ be the kinetic energy of the system, and $$V$$ the potential energy.\begin{align*} T & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \left ( \dot{x}\cos \theta +\dot{\theta }r\right ) ^{2}+\left ( \dot{x}\sin \theta +\dot{r}\right ) ^{2}\right ) \\ V & =\frac{1}{2}kx^{2}+\frac{1}{2}k_{p}\left ( r-l_{0}\right ) ^{2}-mgr\cos \theta \end{align*}

Hence\begin{align*} L & =T-V\\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \left ( \dot{x}^{2}\cos ^{2}\theta +\dot{\theta }^{2}r^{2}+2\dot{x}\dot{\theta }r\cos \theta \right ) +\left ( \dot{x}^{2}\sin ^{2}\theta +\dot{r}^{2}+2\dot{r}\dot{x}\sin \theta \right ) \right ) \\ & -\left ( \frac{1}{2}kx^{2}+\frac{1}{2}k_{p}\left ( r-l_{0}\right ) ^{2}-mgr\cos \theta \right ) \\ & \\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}^{2}+\dot{\theta }^{2}r^{2}+2\dot{x}\dot{\theta }r\cos \theta +\dot{r}^{2}+2\dot{r}\dot{x}\sin \theta \right ) -\frac{1}{2}kx^{2}-\frac{1}{2}k_{p}\left ( r-l_{0}\right ) ^{2}+mgr\cos \theta \end{align*}

For $$x$$ we obtain\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x} & =Q_{x}\\ kx+m\sin \theta \left ( -r\dot{\theta }^{2}+\ddot{r}\right ) +\left ( m+M\right ) \ddot{x}+m\cos \theta \left ( 2\dot{r}\dot{\theta }+r\ddot{\theta }\right ) & =Q_{x} \end{align*}

To ﬁnd $$Q_{x}$$ we make small $$\delta x$$ displacement and ﬁnd the virtual work done. Hence we obtain$\delta W=F\delta x$ Hence $$Q_{x}=F\left ( t\right )$$ and EQM for $$x$$ becomes$kx+m\sin \theta \left ( -r\dot{\theta }^{2}+\ddot{r}\right ) +\left ( m+M\right ) \ddot{x}+m\cos \theta \left ( 2\dot{r}\dot{\theta }+r\ddot{\theta }\right ) =F$ Hence$$\ddot{x}=\frac{F-kx-m\sin \theta \left ( -r\dot{\theta }^{2}+\ddot{r}\right ) -m\cos \theta \left ( 2\dot{r}\dot{\theta }+r\ddot{\theta }\right ) }{\left ( m+M\right ) }\tag{1}$$ For $$r$$ we obtain\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{r}}-\frac{\partial L}{\partial r} & =Q_{r}\\ -k_{p}l_{0}-gm\cos \theta +r\left ( k_{p}-m\dot{\theta }^{2}\right ) +m\ddot{r}+m\ddot{x}\sin \theta & =Q_{r} \end{align*}

To ﬁnd $$Q_{r}$$ we make small $$\delta r$$ displacement and ﬁnd the virtual work done. Hence we obtain $$Q_{r}=0$$ and EQM for $$r$$ becomes$-k_{p}l_{0}-gm\cos \theta +r\left ( k_{p}-m\dot{\theta }^{2}\right ) +m\ddot{r}+m\ddot{x}\sin \theta =0$ Hence$$\ddot{r}=\frac{k_{p}l_{0}+gm\cos \theta -r\left ( k_{p}-m\dot{\theta }^{2}\right ) -m\ddot{x}\sin \theta }{m}\tag{2}$$ For $$\theta$$ we obtain\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta } & =Q_{\theta }\\ mr\left ( g\sin \theta +2\dot{r}\dot{\theta }+\cos \theta \ddot{x}+r\ddot{\theta }\right ) & =Q_{\theta } \end{align*}

To ﬁnd $$Q_{\theta }$$ we make small $$\delta \theta$$ displacement and ﬁnd the virtual work done. Hence we obtain $$Q_{\theta }=0$$ and EQM for $$\theta$$ becomes$mr\left ( g\sin \theta +2\dot{r}\dot{\theta }+\cos \theta \ddot{x}+r\ddot{\theta }\right ) =0$ Hence $$\ddot{\theta }=\frac{-g\sin \theta -2\dot{r}\dot{\theta }-\ddot{x}\cos \theta }{r}\tag{3}$$ To be able to convert to ﬁrst order form, we need to decouple the above equations. This results in\begin{align*} \ddot{x} & =\frac{F\left ( t\right ) }{M}-\frac{k}{M}x-\frac{k_{p}}{M}\left ( r-l_{0}\right ) \sin \theta \\ \ddot{r} & =g\cos \theta -\frac{k_{p}}{m}\left ( r-l_{0}\right ) -\frac{F\left ( t\right ) }{M}\sin \theta +\frac{k}{M}x\sin \theta +\frac{k_{p}}{M}\left ( r-l_{0}\right ) \sin ^{2}\theta +\dot{\theta }^{2}r\\ \ddot{\theta } & =-\frac{g}{r}\sin \theta -\frac{2\dot{r}}{r}\dot{\theta }-\frac{F\left ( t\right ) }{rM}\cos \theta +\frac{k}{rM}x\cos \theta +\frac{k_{p}}{rM}\left ( r-l_{0}\right ) \sin \theta \cos \theta \end{align*}

##### Convert to state space form

Using the decoupled ODE’s above, and introducing 6 state variables $$x_{1},x_{2},x_{3,}\,x_{4},x_{5},x_{6}$$ we obtain\begin{align*} \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\\ x_{6}\end{pmatrix} & =\begin{pmatrix} x\\ \theta \\ r\\ x^{\prime }\\ \theta ^{\prime }\\ r^{\prime }\end{pmatrix} \overset{\frac{d}{dt}}{\rightarrow }\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\\ \dot{x}_{5}\\ \dot{x}_{6}\end{pmatrix} =\begin{pmatrix} x^{\prime }\\ \theta ^{\prime }\\ r^{\prime }\\ x^{\prime \prime }\\ \theta ^{\prime \prime }\\ r^{\prime \prime }\end{pmatrix} \\\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\\ \dot{x}_{5}\\ \dot{x}_{6}\end{pmatrix} & =\begin{pmatrix} x_{4}\\ x_{5}\\ x_{6}\\ \frac{F\left ( t\right ) }{M}-\frac{k}{M}x-\frac{k_{p}}{M}\left ( r-l_{0}\right ) \sin \theta \\ -\frac{g}{r}\sin \theta -\frac{2r^{\prime }}{r}\theta ^{\prime }-\frac{F\left ( t\right ) }{rM}\cos \theta +\frac{k}{rM}x\cos \theta +\frac{k_{p}}{rM}\left ( r-l_{0}\right ) \sin \theta \cos \theta \\ g\cos \theta -\frac{k_{p}}{m}\left ( r-l_{0}\right ) -\frac{F\left ( t\right ) }{M}\sin \theta +\frac{k}{M}x\sin \theta +\frac{k_{p}}{M}\left ( r-l_{0}\right ) \sin ^{2}\theta +\dot{\theta }^{2}r \end{pmatrix} \\\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\\ \dot{x}_{5}\\ \dot{x}_{6}\end{pmatrix} & =\begin{pmatrix} x_{4}\\ x_{5}\\ x_{6}\\ \frac{F\left ( t\right ) }{M}-\frac{k}{M}x_{1}-\frac{k_{p}}{M}\left ( x_{3}-l_{0}\right ) \sin x_{2}\\ -\frac{g}{r}\sin x_{2}-\frac{2x_{6}}{x_{3}}x_{5}-\frac{F\left ( t\right ) }{x_{3}M}\cos x_{2}+\frac{k}{x_{3}M}x_{1}\cos x_{2}+\frac{k_{p}}{x_{3}M}\left ( x_{3}-l_{0}\right ) \sin x_{2}\cos x_{2}\\ g\cos x_{2}-\frac{k_{p}}{m}\left ( x_{3}-l_{0}\right ) -\frac{F\left ( t\right ) }{M}\sin x_{2}+\frac{k}{M}x_{1}\sin x_{2}+\frac{k_{p}}{M}\left ( x_{3}-l_{0}\right ) \sin ^{2}x_{2}+x_{5}^{2}x_{3}\end{pmatrix} \end{align*}

#### 3.2.5problem 4

##### Velocity Diagram

Only the velocity diagram is needed for the Lagrangian method. Generalized coordinates are: $$x,\theta$$

Let $$L$$ be the Lagrangian, and let $$T$$ be the kinetic energy of the system, and $$V$$ the potential energy.\begin{align*} T & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \left ( \dot{x}-L\dot{\theta }\right ) ^{2}+\left ( L\dot{\theta }\cos \theta \right ) ^{2}\right ) \\ V & =Mgx+mg\left ( L\cos \theta +x\right ) \end{align*}

Hence\begin{align*} L & =T-V\\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}^{2}+L^{2}\dot{\theta }^{2}-2\dot{x}L\dot{\theta }+L^{2}\dot{\theta }^{2}\cos ^{2}\theta \right ) -Mgx-mg\left ( L\cos \theta +x\right ) \end{align*}

For $$x$$ we obtain\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x} & =Q_{x}\\ g\left ( m+M\right ) +\left ( m+M\right ) \ddot{x}-Lm\ddot{\theta } & =Q_{x} \end{align*}

To ﬁnd $$Q_{x}$$ we make small $$\delta x$$ displacement and ﬁnd the virtual work done. Hence we obtain$\delta W=F\delta x$ Therefore $$Q_{x}=F$$ and the EQM for $$x$$ becomes$g\left ( m+M\right ) +\left ( m+M\right ) \ddot{x}-Lm\ddot{\theta }=F$ Hence $$\ddot{x}=\frac{F-g\left ( m+M\right ) +Lm\ddot{\theta }}{m+M}\tag{1}$$

For $$\theta$$ we have

\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta } & =Q_{\theta }\\ Lm\left ( g\sin \theta +L\dot{\theta }^{2}\cos \theta \sin \theta +\ddot{x}-L\ddot{\theta }\left ( 1+\cos ^{2}\theta \right ) \right ) & =Q_{\theta } \end{align*}

To ﬁnd $$Q_{\theta }$$ we make small $$\delta \theta$$ displacement and ﬁnd the virtual work done. We ﬁnd that $$Q_{\theta }=0$$ hence the EQM for $$\theta$$ becomes

\begin{align*} Lm\left ( g\sin \theta +L\dot{\theta }^{2}\cos \theta \sin \theta +\ddot{x}-L\ddot{\theta }\left ( 1+\cos ^{2}\theta \right ) \right ) & =0\\ g\sin \theta +L\dot{\theta }^{2}\cos \theta \sin \theta +\ddot{x}-L\ddot{\theta }\left ( 1+\cos ^{2}\theta \right ) & =0 \end{align*}

Hence

$$\ddot{\theta }=\frac{g\sin \theta +L\dot{\theta }^{2}\cos \theta \sin \theta +\ddot{x}}{L\left ( 1+\cos ^{2}\theta \right ) }\tag{2}$$

To convert to ﬁrst form, Eqs. (1) and (2) are decoupled resulting in

\begin{align*} \ddot{x} & =\frac{\left ( F-g\left ( m+M\right ) \right ) \left ( 1+\cos ^{2}\theta \right ) +mg\sin \theta +Lm\dot{\theta }^{2}\cos \theta \sin \theta }{M+\left ( M+m\right ) \cos ^{2}\theta }\\ \ddot{\theta } & =\frac{F-g\left ( m+M\right ) +g\left ( m+M\right ) \sin \theta +L\left ( m+M\right ) \dot{\theta }^{2}\cos \theta \sin \theta }{L\left ( M+\left ( M+m\right ) \cos ^{2}\theta \right ) } \end{align*}

##### Convert to state space form

Using the decoupled ODE’s above, and introducing 4 state variables $$x_{1},x_{2},x_{3,}\,x_{4}$$ we obtain\begin{align*} \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} & =\begin{pmatrix} x\\ \theta \\ x^{\prime }\\ \theta ^{\prime }\end{pmatrix} \overset{\frac{d}{dt}}{\rightarrow }\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} =\begin{pmatrix} x^{\prime }\\ \theta ^{\prime }\\ x^{\prime \prime }\\ \theta ^{\prime \prime }\end{pmatrix} \\\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} & =\begin{pmatrix} x_{3}\\ x_{4}\\ \frac{\left ( F-g\left ( m+M\right ) \right ) \left ( 1+\cos ^{2}\theta \right ) +mg\sin \theta +Lm\dot{\theta }^{2}\cos \theta \sin \theta }{M+\left ( M+m\right ) \cos ^{2}\theta }\\ \frac{F-g\left ( m+M\right ) +g\left ( m+M\right ) \sin \theta +L\left ( m+M\right ) \dot{\theta }^{2}\cos \theta \sin \theta }{L\left ( M+\left ( M+m\right ) \cos ^{2}\theta \right ) }\end{pmatrix} \\ & =\begin{pmatrix} x_{3}\\ x_{4}\\ \frac{\left ( F-g\left ( m+M\right ) \right ) \left ( 1+\cos ^{2}x_{2}\right ) +mg\sin x_{2}+Lmx_{4}^{2}\cos x_{2}\sin x_{2}}{M+\left ( M+m\right ) \cos ^{2}x_{2}}\\ \frac{F-g\left ( m+M\right ) +g\left ( m+M\right ) \sin x_{2}+L\left ( m+M\right ) x_{4}^{2}\cos x_{2}\sin x_{2}}{L\left ( M+\left ( M+m\right ) \cos ^{2}x_{2}\right ) }\end{pmatrix} \end{align*}

#### 3.2.6Solution problem 5

Only the velocity diagram is needed for the Lagrangian method.

Let $$L$$ be the Lagrangian, and let $$T$$ be the kinetic energy of the system, and $$V$$ the potential energy.\begin{align*} T & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}+R\dot{\theta }\right ) ^{2}+\frac{1}{2}I_{g}\dot{\theta }^{2}\\ V & =\frac{1}{2}k\left ( R\theta \right ) ^{2} \end{align*}

Hence\begin{align*} L & =T-V\\ & =\left [ \frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}+R\dot{\theta }\right ) ^{2}+\frac{1}{2}I_{g}\dot{\theta }^{2}\right ] -\frac{1}{2}k\left ( R\theta \right ) ^{2}\\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}^{2}+R^{2}\dot{\theta }^{2}+2\dot{x}\dot{\theta }R\right ) +\frac{1}{2}I_{g}\dot{\theta }^{2}-\frac{1}{2}kR^{2}\theta ^{2} \end{align*}

and\begin{align*} \frac{\partial L}{\partial \dot{x}} & =M\dot{x}+m\dot{x}+m\dot{\theta }R\\ \frac{\partial L}{\partial x} & =0\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} & =M\ddot{x}+m\ddot{x}+m\ddot{\theta }R \end{align*}

Hence, for $$x$$, the EQM is$M\ddot{x}+m\ddot{x}+m\ddot{\theta }R=Q_{x}$ Where $$Q_{x}$$ is the generalized force which is\begin{align*} \delta W & =\frac{F\delta x+\left ( kR\theta \right ) \delta x+\left ( bR\dot{\theta }\right ) \delta x}{\delta x}\\ & =F+bR\dot{\theta }+kR\theta \end{align*}

hence\begin{align*} M\ddot{x}+m\ddot{x}+m\ddot{\theta }R & =F\left ( t\right ) +bR\dot{\theta }+kR\theta \\ \ddot{x} & =\frac{F\left ( t\right ) +bR\dot{\theta }+kR\theta -m\ddot{\theta }R}{M+m} \end{align*}

Since $$I_{g}=\frac{mR^{2}}{2}$$, then the above can be written as$\ddot{x}=\frac{RF\left ( t\right ) +bR^{2}\dot{\theta }+kR^{2}\theta -2I_{g}\ddot{\theta }}{R\left ( M+m\right ) }$ For $$\theta$$, the EQM is\begin{align*} \frac{\partial L}{\partial \dot{\theta }} & =m\left ( R^{2}\dot{\theta }+\dot{x}R\right ) +I_{g}\dot{\theta }\\ \frac{\partial L}{\partial \theta } & =-kR^{2}\theta \\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} & =m\left ( R^{2}\ddot{\theta }+\ddot{x}R\right ) +I_{g}\ddot{\theta } \end{align*}

Hence$m\left ( R^{2}\ddot{\theta }+\ddot{x}R\right ) +I_{g}\ddot{\theta }+kR^{2}\theta =F_{\theta }$ In this case, $$\delta W=-\frac{\left ( bR\dot{\theta }\right ) R\delta \theta }{\delta \theta }$$, hence $$F_{\theta }=-bR^{2}\dot{\theta }$$, therefore, the EQM is\begin{align*} m\left ( R^{2}\ddot{\theta }+\ddot{x}R\right ) +I_{g}\ddot{\theta }+kR^{2}\theta & =-bR^{2}\dot{\theta }\\ \ddot{\theta } & =\frac{-bR^{2}\dot{\theta }-kR^{2}\theta -m\ddot{x}R}{I_{g}+mR^{2}}\\ & =-\frac{bR^{2}\dot{\theta }+kR^{2}\theta +m\ddot{x}R}{I_{o}} \end{align*}

By decoupling the 2 equations of motion we obtain\begin{align*} x^{\prime \prime } & =\frac{I_{o}F\left ( t\right ) +kI_{o}R\theta +bI_{o}R\theta ^{\prime }+I_{g}\left [ kR\theta +bR\theta ^{\prime }\right ] }{MI_{o}-I_{g}m}\\ \theta ^{\prime \prime } & =\frac{-mRF\left ( t\right ) -mkR^{2}\theta -mbR^{2}\theta ^{\prime }-MkR^{2}\theta -MbR^{2}\theta ^{\prime }}{MI_{o}-I_{g}m} \end{align*}

##### Convert to state space form

Using the decoupled ODE’s above, and introducing 4 state variables $$x_{1},x_{2},x_{3,}\,x_{4}$$ we obtain\begin{align*} \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} & =\begin{pmatrix} x\\ \theta \\ x^{\prime }\\ \theta ^{\prime }\end{pmatrix} \overset{\frac{d}{dt}}{\rightarrow }\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} =\begin{pmatrix} x^{\prime }\\ \theta ^{\prime }\\ x^{\prime \prime }\\ \theta ^{\prime \prime }\end{pmatrix} \\\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \dot{x}_{4}\end{pmatrix} & =\begin{pmatrix} x_{3}\\ x_{4}\\ \frac{I_{o}F\left ( t\right ) +kI_{o}R\theta +bI_{o}R\theta ^{\prime }+I_{g}\left [ kR\theta +bR\theta ^{\prime }\right ] }{MI_{o}-I_{g}m}\\ \frac{-mRF\left ( t\right ) -mkR^{2}\theta -mbR^{2}\theta ^{\prime }-MkR^{2}\theta -MbR^{2}\theta ^{\prime }}{MI_{o}-I_{g}m}\end{pmatrix} \\ & =\begin{pmatrix} x_{3}\\ x_{4}\\ \frac{I_{o}F\left ( t\right ) +kI_{o}Rx_{2}+bI_{o}Rx_{4}+I_{g}\left [ kRx_{2}+bRx_{4}\right ] }{MI_{o}-I_{g}m}\\ \frac{-mRF\left ( t\right ) -mkR^{2}x_{2}-mbR^{2}x_{4}-MkR^{2}x_{2}-MbR^{2}x_{4}}{MI_{o}-I_{g}m}\end{pmatrix} \end{align*}