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