4.2 Second HW set, Solve the first set using Lagrangian

solve the first practice problems using Lagrangian approach.

4.2.1 problem 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

\begin{gather} M\ddot {x}+m\ddot {x}+mL\ddot {\theta }\cos \theta -mL\dot {\theta }^{2}\sin \theta +kx=F\nonumber \\ \ddot {x}=\frac {F-mL\ddot {\theta }\cos \theta +mL\dot {\theta }^{2}\sin \theta -kx}{M+m}\tag {1}\end{gather}

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}

4.2.2 Decouple 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*}

4.2.3 problem 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

\begin{gather} \theta ^{\prime \prime }I_{o}\left ( M+m\right ) =\left ( M+m\right ) mag\sin \theta -ma\left [ F\left ( t\right ) -kx-ma\theta ^{\prime \prime }\cos \theta +ma\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \right ] \cos \theta \nonumber \\ =\left ( M+m\right ) mag\sin \theta -maF\left ( t\right ) \cos \theta +makx\cos \theta +m^{2}a^{2}\theta ^{\prime \prime }\cos ^{2}\theta -m^{2}a^{2}\left ( \theta ^{\prime }\right ) ^{2}\sin \theta \cos \theta \nonumber \\ \theta ^{\prime \prime }\left [ I_{o}\left ( M+m\right ) -m^{2}a^{2}\cos ^{2}\theta \right ] =\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 \nonumber \\ \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 }\tag {4}\end{gather}

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*}

4.2.4 Solution 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

\begin{equation} \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}\end{equation}

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

\begin{equation} \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}\end{equation}

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

\begin{equation} \ddot {\theta }=\frac {-g\sin \theta -2\dot {r}\dot {\theta }-\ddot {x}\cos \theta }{r}\tag {3}\end{equation}

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*}

4.2.5 problem 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

\begin{equation} \ddot {x}=\frac {F-g\left ( m+M\right ) +Lm\ddot {\theta }}{m+M}\tag {1}\end{equation}

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

\begin{equation} \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}\end{equation}

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*}

4.2.6 Solution 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*}