## 5 Dynamic Analysis  
### Dynamic Model  
The proposed mechanism has one degree of freedom and can thus be modeled using a single equivalent mass \(m^{*}\) that is excited via a single equivalent force \(F^{*}\). The equivalent mass and force clearly depend on the mechanism configuration (\(x\)) and on its main parameters.  
The equations of motion are obtained by recalling that the power equals the derivative of the kinetic energy:  
\[P=\frac{dE_{k}}{dt} \tag{21}\]  
Since friction between the members of the mechanism and their potential energy can be neglected. The kinetic energy of the equivalent mass equals by definition the kinetic energy of all the members of the mechanism:  
\[E_{k}=\frac{1}{2}m^{*}\dot{x}^{2}=\frac{1}{2}\bigg{(}m\dot{x}^{2}+m_{r}\Big{(} \dot{x}_{Gr}^{2}+\dot{y}_{Gr}^{2}\Big{)}+I_{r}\dot{\alpha}^{2}+m_{c}\Big{(}\dot {x}_{Gc}^{2}+\dot{y}_{Gc}^{2}\Big{)}+I_{c}\dot{\theta}^{2}\bigg{)} \tag{22}\]  
where \(\dot{x}\) is the velocity of the equivalent mass (equal to the one of the slider); \(m\) is the mass of the slider (included the spindle and the grinding wheel); \(m_{r}\), \(I_{r}\), \(\dot{x}_{Gr}\), \(\dot{y}_{Gr}\) are the mass, barycentric moment of inertia, barycenter velocity along the \(x\) axis, and barycenter velocity along the \(y\) axis of the rocker arm link, respectively; \(m_{c}\), \(I_{c}\), \(\dot{x}_{Gr}\), \(\dot{y}_{Gr}\) are the mass, barycentric moment of inertia, barycenter velocity along the \(x\) axis, and barycenter velocity along the \(y\) axis of the crank, respectively. The coordinates of the centers of mass of the rocker arm link and of the crank are (respectively):  
\[\left\{\begin{array}{c}x_{Gr}\\ y_{Gr}\end{array}\right\}=\left\{\begin{array}{c}x+\frac{b_{s}}{2}\cos( \alpha)\\ \frac{b_{s}}{2}\sin(\alpha)\end{array}\right\},\ \ \left\{\begin{array}{c}x_{Gc}\\ y_{Gc}\end{array}\right\}=\left\{\begin{array}{c}I_{1}+l_{2}-\frac{l_{1}}{2 }\cos(\theta)\\ \frac{l_{1}}{2}\sin(\theta)\end{array}\right\} \tag{23}\]  
The corresponding velocities are:  
\[\left\{\begin{array}{c}\dot{x}_{Gr}\\ \dot{y}_{Gr}\end{array}\right\}=\left\{\begin{array}{c}\dot{x}-\frac{b_{s}}{ 2}\sin(\alpha)\dot{\alpha}\\ \frac{l_{2}}{2}\cos(\alpha)\dot{\alpha}\end{array}\right\},\ \ \left\{\begin{array}{c}\dot{x}_{Gc}\\ \dot{y}_{Gc}\end{array}\right\}=\left\{\begin{array}{c}\frac{l_{1}}{2}\sin( \theta)\dot{\theta}\\ \frac{l_{2}}{2}\cos(\theta)\dot{\theta}\end{array}\right\} \tag{24}\]  
By introducing (24) and the speed ratios of (11), the final form of (22) is:  
\[\begin{array}{c}\frac{1}{2}m^{*}\dot{x}^{2}=\frac{1}{2}\bigg{(}m+m_{r}\bigg{(} 1+\frac{l_{2}^{2}}{4}\tau_{ax}^{2}-l_{2}\sin(\alpha)\tau_{ax}\bigg{)}+I_{r} \tau_{ax}^{2}+m_{c}\bigg{(}\frac{l_{1}^{2}}{4}\tau_{bx}^{2}\bigg{)}\\ +I_{c}\tau_{gx}^{2}\bigg{)}\dot{x}^{2}\end{array} \tag{25}\]  
The power \(P\) of \(F^{*}\) can be calculated from the deburring force and the springs forces:  
\[P=F^{*}\dot{x}=-(k_{1}s_{1})\dot{s}_{1}-(k_{2}s_{2})\dot{s}_{2}-(k_{3}x)\dot{ x}-\big{(}c_{eq}\dot{x}\big{)}\dot{x}+F_{b}\dot{x} \tag{26}\]  
in which the force of a damper to be connected in parallel to spring \(k_{3}\) (with damping \(c_{eq}\)) is introduced in order to avoid undesired vibrations.  
The equation of motion is found by differentiating (25) and putting it equal to (26):  
\[\begin{array}{c}m^{*}\ddot{x}+\frac{1}{2}\bigg{(}\bigg{(}2I_{r}+\frac{l_{2} ^{2}}{2}m_{r}\bigg{)}\tau_{ax}\dot{\tau}_{ax}+\bigg{(}2I_{c}+\frac{l_{2}^{2}}{ 2}m_{c}\bigg{)}\tau_{gx}\dot{\tau}_{gx}\\ -m_{r}I_{2}\big{(}\tau_{ax}\ \alpha\.\cos(\alpha)+\ \dot{\tau}_{ax}\sin(\alpha) \big{)}\dot{x}^{2}\ =F^{*}\dot{x}\end{array} \tag{27}\]  
The differential Equation (27) is implemented and solved numerically in MATLAB for different working scenarios. The simulation results and the simulation parameters used are presented in Section 5.