Loading web-font TeX/Math/Italic

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).