2.1 my typed solution of EME 121 midterm

This is my EME 121 midterm post-mortem write up.

2.1.1 Problem 1

Cart moves on top of circle as shown, with pendulum attached. Find equations of motion. This is 2 DOF, using \(\theta \) and \(\phi \) for generalized coordinates.

2.1.2 Velocity, acceleration and force diagrams

For the mass \(m\) we use  \(V_{p}=V_{o}+\omega \times r+V_{rel}\). The important thing to see is that \(\omega \) of the frame of reference for \(m\) is relative to the inertial frame always. Hence \(\omega =\dot {\phi }+\dot {\theta }\). In this problem, we see that \(V_{rel}=0\) for mass \(m\) since the length \(L\) do not change. Therefore, for \(m\) we have

\begin{align*} V_{p} & =V_{o}+\omega \times r+V_{rel}\\ & =R\dot {\theta }+L\left ( \dot {\phi }+\dot {\theta }\right ) \end{align*}

For the acceleration, we have

\[ a_{p}=a_{0}+\dot {\omega }\times r+\omega \times \left ( \omega \times r\right ) +2\omega \times V_{rel}+a_{rel}\]

For the mass \(m\), \(V_{rel}\) and \(a_{rel}\) are zero, since the mass \(m\) is on a fixed rod. So we have

\begin{align*} a_{p} & =a_{0}+\left ( \ddot {\phi }+\ddot {\theta }\right ) \times r+\left ( \dot {\phi }+\dot {\theta }\right ) \times \left ( \left ( \dot {\phi }+\dot {\theta }\right ) \times r\right ) \\ & =R\ddot {\theta }+L\left ( \ddot {\phi }+\ddot {\theta }\right ) +\left ( \dot {\phi }+\dot {\theta }\right ) \left ( \dot {\phi }+\dot {\theta }\right ) L\\ & =R\ddot {\theta }+L\left ( \ddot {\phi }+\ddot {\theta }\right ) +L\dot {\phi }^{2}+L\dot {\theta }^{2}+2L\dot {\phi }\dot {\theta }\\ & =R\ddot {\theta }+L\left ( \ddot {\phi }+\ddot {\theta }\right ) +L\left ( \dot {\phi }^{2}+\dot {\theta }^{2}\right ) +2L\dot {\phi }\dot {\theta }\end{align*}

So, in diagram, this is how it looks

The unknowns are \(\theta ,\phi ,T,N\), hence we need 4 equations.

2.1.3 Solution using \(F=ma\)

For pendulum mass \(m\), resolving forces radially we obtain

\begin{align} mg\cos \left ( \theta +\phi \right ) -T & =m\left ( L\left ( \dot {\phi }^{2}+\dot {\theta }^{2}\right ) +2L\dot {\phi }\dot {\theta }+R\dot {\theta }^{2}\cos \phi -R\ddot {\theta }\sin \phi \right ) \nonumber \\ T & =m\left [ g\cos \left ( \theta +\phi \right ) -L\left ( \dot {\phi }^{2}+\dot {\theta }^{2}\right ) -2L\dot {\phi }\dot {\theta }-R\dot {\theta }^{2}\cos \phi +R\ddot {\theta }\sin \phi \right ] \tag {1}\end{align}

and resolving forces tangentially gives

\begin{align} mg\sin \left ( \theta +\phi \right ) & =m\left ( L\left ( \ddot {\phi }+\ddot {\theta }\right ) +R\dot {\theta }^{2}\sin \phi +R\ddot {\theta }\cos \phi \right ) \nonumber \\ \ddot {\phi } & =\frac {g}{L}\sin \left ( \theta +\phi \right ) -\frac {R}{L}\left ( \dot {\theta }^{2}\sin \phi +\ddot {\theta }\cos \phi \right ) -L\ddot {\theta }\tag {2}\end{align}

For the cart \(M\,\), resolving forces radially we obtain

\begin{equation} N-T\cos \phi -Mg\cos \theta =-MR\dot {\theta }^{2}\tag {3}\end{equation}

and resolving forces tangentially gives

\begin{equation} Mg\sin \theta -T\sin \phi =MR\ddot {\theta }\tag {4}\end{equation}

We now have 4 equations with 4 unknowns. Using Mathematica the solution is

Convert to first form

Let

\[\begin {pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end {pmatrix} =\begin {pmatrix} \theta \\ \phi \\ \dot {\theta }\\ \dot {\phi }\end {pmatrix} \rightarrow \begin {pmatrix} \dot {x}_{1}\\ \dot {x}_{2}\\ \dot {x}_{3}\\ \dot {x}_{4}\end {pmatrix} =\begin {pmatrix} x_{3}\\ x_{4}\\ \ddot {\theta }\\ \ddot {\phi }\end {pmatrix} \]

Now since the equations are decoupled, the rest follows.

2.1.4 Problem 2

The disk rolls with no slip, spring has no friction on the side with the disk slot. Find equations of motion.

2 DOF, using \(x\) and \(\theta \) for generalized coordinates.

Velocity, acceleration and force diagrams

Solution using \(F=ma\)

There are 5 unknowns \(\theta ,x,T,N,F_{f}\), hence we need 5 equations.

For mass \(m\), resolving forces along \(x\) direction, positive upward

\begin{equation} -kx-mg\cos \theta =m\left ( \ddot {x}-x\dot {\theta }^{2}+R\ddot {\theta }\sin \theta \right ) \tag {1}\end{equation}

Resolving tangential to \(x\) direction gives

\begin{equation} mg\sin \theta -T=m\left ( R\ddot {\theta }\cos \theta +x\ddot {\theta }+2\dot {x}\dot {\theta }\right ) \tag {2}\end{equation}

For mass \(M\), resolving horizontally, positive to the right results in

\begin{equation} -F_{f}+T\cos \theta +kx\sin \theta =M\left ( R\ddot {\theta }\right ) \tag {3}\end{equation}

Resolving vertically, positive upwards gives

\begin{equation} -Mg+N-T\sin \theta +kx\cos \theta =0 \tag {4}\end{equation}

Applying moments around c.g. for disk results in

\begin{equation} \tau +F_{f}R+Tx=I\ddot {\theta } \tag {5}\end{equation}

Now we have 5 equations and 5 unknowns. Solving for \(\ddot {\theta },\ddot {x}\) gives

\begin{align*} \theta ^{\prime \prime }\left ( t\right ) & =\frac {\tau +gmR\cos \theta \sin \theta +\left ( gm+Rk\right ) x\sin \theta -2m\left ( R\cos \theta +x\right ) \dot {x}\dot {\theta }}{I+MR^{2}+mR^{2}\cos ^{2}\theta +mx\left ( 2R\cos \theta +x\right ) }\\ x^{\prime \prime }\left ( t\right ) & =-g\cos \theta +x\left ( \dot {\theta }^{2}-\frac {k}{m}\right ) -\frac {R\sin \theta \left ( \tau +kRx\sin \theta +m\left ( R\cos \theta +x\right ) \left ( g\sin \theta -2\dot {x}\dot {\theta }\right ) \right ) }{I+MR^{2}+mR^{2}\cos ^{2}\theta +mx\left ( 2R\cos \theta +x\right ) }\end{align*}

2.1.5 Convert to first form

Let

\[\begin {pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end {pmatrix} =\begin {pmatrix} \theta \\ x\\ \dot {\theta }\\ \dot {x}\end {pmatrix} \rightarrow \begin {pmatrix} \dot {x}_{1}\\ \dot {x}_{2}\\ \dot {x}_{3}\\ \dot {x}_{4}\end {pmatrix} =\begin {pmatrix} x_{3}\\ x_{4}\\ \ddot {\theta }\\ \ddot {x}\end {pmatrix} \]

Hence

\begin{align*}\begin {pmatrix} \dot {x}_{1}\\ \dot {x}_{2}\\ \dot {x}_{3}\\ \dot {x}_{4}\end {pmatrix} & =\begin {pmatrix} x_{3}\\ x_{4}\\ \frac {\tau +gmR\cos \theta \sin \theta +\left ( gm+Rk\right ) x\sin \theta -2m\left ( R\cos \theta +x\right ) \dot {x}\dot {\theta }}{I+MR^{2}+mR^{2}\cos ^{2}\theta +mx\left ( 2R\cos \theta +x\right ) }\\ -g\cos \theta +x\left ( \dot {\theta }^{2}-\frac {k}{m}\right ) -\frac {R\sin \theta \left ( \tau +kRx\sin \theta +m\left ( R\cos \theta +x\right ) \left ( g\sin \theta -2\dot {x}\dot {\theta }\right ) \right ) }{I+MR^{2}+mR^{2}\cos ^{2}\theta +mx\left ( 2R\cos \theta +x\right ) }\end {pmatrix} \\ & =\begin {pmatrix} x_{3}\\ x_{4}\\ \frac {\tau +gmR\cos x_{1}\sin x_{1}+\left ( gm+Rk\right ) x_{2}\sin x_{1}-2m\left ( R\cos x_{1}+x_{2}\right ) x_{4}x_{3}}{I+MR^{2}+mR^{2}\cos ^{2}x_{1}+mx\left ( 2R\cos x_{1}+x_{2}\right ) }\\ -g\cos x_{1}+x_{2}\left ( x_{3}^{2}-\frac {k}{m}\right ) -\frac {R\sin x_{1}\left ( \tau +kRx_{2}\sin x_{1}+m\left ( R\cos x_{1}+x_{2}\right ) \left ( g\sin x_{1}-2x_{4}x_{3}\right ) \right ) }{I+MR^{2}+mR^{2}\cos ^{2}x_{1}+mx_{2}\left ( 2R\cos x_{1}+x_{2}\right ) }\end {pmatrix} \end{align*}

2.1.6 Problem 3

The kinetic energy is

\begin{align*} T & =\frac {1}{2}M\left ( R\dot {\theta }\right ) ^{2}+\frac {1}{2}I\dot {\theta }^{2}+\frac {1}{2}m\left [ \left ( \dot {x}+R\dot {\theta }\sin \theta \right ) ^{2}+\left ( R\dot {\theta }\cos \theta +x\dot {\theta }\right ) ^{2}\right ] \\ & =\frac {1}{2}M\left ( R\dot {\theta }\right ) ^{2}+\frac {1}{2}I\dot {\theta }^{2}+\frac {1}{2}m\left [ \dot {x}^{2}+2R\dot {x}\dot {\theta }\sin \theta +R^{2}\dot {\theta }^{2}+x^{2}\dot {\theta }^{2}+2x\dot {\theta }^{2}R\cos \theta \right ] \\ & =\frac {1}{2}M\left ( R\dot {\theta }\right ) ^{2}+\frac {1}{2}I\dot {\theta }^{2}+\frac {1}{2}m\dot {x}^{2}+mR\dot {x}\dot {\theta }\sin \theta +\frac {1}{2}m\dot {\theta }^{2}\left ( R^{2}+x^{2}+2xR\cos \theta \right ) \end{align*}

and the potential energy is

\[ V=\frac {1}{2}kx^{2}+mgx\cos \theta \]

Hence

\begin{align*} L & =T-V\\ & =\frac {1}{2}M\left ( R\dot {\theta }\right ) ^{2}+\frac {1}{2}I\dot {\theta }^{2}+\frac {1}{2}m\dot {x}^{2}+mR\dot {x}\dot {\theta }\sin \theta +\frac {1}{2}m\dot {\theta }^{2}\left ( R^{2}+x^{2}+2xR\cos \theta \right ) -\frac {1}{2}kx^{2}-mgx\cos \theta \end{align*}

For \(x\) we have

\begin{align*} \frac {\partial L}{\partial \dot {x}} & =m\dot {x}+mR\dot {\theta }\sin \theta \\ \frac {d}{dt}\frac {\partial L}{\partial \dot {x}} & =m\ddot {x}+mR\ddot {\theta }\sin \theta +mR\dot {\theta }^{2}\cos \theta \\ \frac {\partial L}{\partial x} & =m\dot {\theta }^{2}\left ( x+R\cos \theta \right ) -kx-mg\cos \theta \end{align*}

Hence equation of motion is

\begin{align*} \frac {d}{dt}\frac {\partial L}{\partial \dot {x}}-\frac {\partial L}{\partial x} & =Q_{x}\\ m\ddot {x}+mR\ddot {\theta }\sin \theta +mR\dot {\theta }^{2}\cos \theta -m\dot {\theta }^{2}\left ( x+R\cos \theta \right ) +kx+mg\cos \theta & =0 \end{align*}

Since generalized force is zero for \(x\). Hence from above, we have

\begin{align} \ddot {x} & =-R\ddot {\theta }\sin \theta -R\dot {\theta }^{2}\cos \theta +\dot {\theta }^{2}\left ( x+R\cos \theta \right ) -\frac {k}{m}x-g\cos \theta \nonumber \\ & =-R\ddot {\theta }\sin \theta +\dot {\theta }^{2}x-\frac {k}{m}x-g\cos \theta \tag {1}\end{align}

Now for \(\theta \)

\begin{align*} \frac {\partial L}{\partial \dot {\theta }} & =MR^{2}\dot {\theta }+I\dot {\theta }+mR\dot {x}\sin \theta +m\dot {\theta }\left ( R^{2}+x^{2}+2xR\cos \theta \right ) \\ \frac {d}{dt}\frac {\partial L}{\partial \dot {\theta }} & =MR^{2}\ddot {\theta }+I\ddot {\theta }+mR\ddot {x}\sin \theta +mR\dot {x}\dot {\theta }\cos \theta +m\ddot {\theta }\left ( R^{2}+x^{2}+2xR\cos \theta \right ) +m\dot {\theta }\left ( 2x\dot {x}+2\dot {x}R\cos \theta -2xR\dot {\theta }\sin \theta \right ) \\ & =MR^{2}\ddot {\theta }+I\ddot {\theta }+mR\ddot {x}\sin \theta +mR\dot {x}\dot {\theta }\cos \theta +m\ddot {\theta }R^{2}+m\ddot {\theta }x^{2}+2m\ddot {\theta }xR\cos \theta +2m\dot {\theta }x\dot {x}+2m\dot {\theta }\dot {x}R\cos \theta -2mxR\dot {\theta }^{2}\sin \theta \\ & =MR^{2}\ddot {\theta }+I\ddot {\theta }+mR\ddot {x}\sin \theta +3mR\dot {x}\dot {\theta }\cos \theta +m\ddot {\theta }R^{2}+m\ddot {\theta }x^{2}+2m\ddot {\theta }xR\cos \theta +2m\dot {\theta }x\dot {x}-2mxR\dot {\theta }^{2}\sin \theta \\ \frac {\partial L}{\partial \theta } & =mR\dot {x}\dot {\theta }\cos \theta -m\dot {\theta }^{2}xR\sin \theta +mgx\sin \theta \end{align*}

Hence equation of motion is

\[ \frac {d}{dt}\frac {\partial L}{\partial \dot {\theta }}-\frac {\partial L}{\partial \theta }=Q_{\theta }\]

Hence, since \(Q_{\theta }=\tau \) then

\[ MR^{2}\ddot {\theta }+I\ddot {\theta }+mR\ddot {x}\sin \theta +m\ddot {\theta }R^{2}+m\ddot {\theta }x^{2}+2xm\ddot {\theta }R\cos \theta +2m\dot {\theta }x\dot {x}+2m\dot {\theta }\dot {x}R\cos \theta -m\dot {\theta }^{2}xR\sin \theta +mgx\sin \theta =\tau \]

or

\begin{align} \ddot {\theta }\left ( I+2xmR\cos \theta +MR^{2}\right ) & =\tau -mR\ddot {x}\sin \theta +m\ddot {\theta }R^{2}+m\ddot {\theta }x^{2}+2m\dot {\theta }x\dot {x}+2m\dot {\theta }\dot {x}R\cos \theta -m\dot {\theta }^{2}xR\sin \theta +mgx\sin \theta \nonumber \\ \ddot {\theta } & =\frac {\tau -mR\ddot {x}\sin \theta +m\ddot {\theta }R^{2}+m\ddot {\theta }x^{2}+2m\dot {\theta }x\dot {x}+2m\dot {\theta }\dot {x}R\cos \theta -m\dot {\theta }^{2}xR\sin \theta +mgx\sin \theta }{I+2xmR\cos \theta +MR^{2}} \tag {2}\end{align}

To decouple Eqs. (1) and (2), substitute (1) into (2), and then (2) into (1) to obtain

\begin{align*} \theta ^{\prime \prime }\left ( t\right ) & =\frac {\tau +gmR\cos \theta \sin \theta +\left ( gm+Rk\right ) x\sin \theta -2m\left ( R\cos \theta +x\right ) \dot {x}\dot {\theta }}{I+MR^{2}+mR^{2}\cos ^{2}\theta +mx\left ( 2R\cos \theta +x\right ) }\\ x^{\prime \prime }\left ( t\right ) & =-g\cos \theta +x\left ( \dot {\theta }^{2}-\frac {k}{m}\right ) -\frac {R\sin \theta \left ( \tau +kRx\sin \theta +m\left ( R\cos \theta +x\right ) \left ( g\sin \theta -2\dot {x}\dot {\theta }\right ) \right ) }{I+MR^{2}+mR^{2}\cos ^{2}\theta +mx\left ( 2R\cos \theta +x\right ) }\end{align*}

These are the same equations found in part (2).