My question is related to a previous question, presented in 342655. I am taking the same scenario to ask a different question: In a flywheel and piston inertial system, how do you calculate the forces transmitted to the frame supporting the flywheel shaft?

Scenario: "Imagine a piston in a cylinder, lying down so that the piston moves horizontally. The cylinder is open at both ends (no compression of a gas). For simplicity, lets assume no friction, no sound, no heat effects, no gravity, and that the system is isolated in a vacuum.

The piston has a connecting rod attached to a revolving flywheel. Consequently the piston oscillates back and forth within the cylinder. The motion of the piston resembles simple harmonic motion ; its kinetic energy oscillates across time, between a maximum at the middle of the cylinder, to zero at either end of the cylinder.

For simple harmonic motion (e.g., a weight attached to a spring, and oscillating horizontally on a frictionless table top), it is well known that oscillation of kinetic energy of the weight is counterbalanced exactly by a coincident oscillation of potential energy (e.g., potential energy due to compression of a spring) in such a way that the total energy remains constant at all times:

\begin{align} & K = E \sin^2(wt) \\ & P = E \cos^2(wt) \\ & K + P = E \end{align}

Furthermore, such motion is indefinite (excluding friction, etc...)."


1 Answer 1


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\begin{align*} &x=\cos \left( \varphi \right) r_{{1}}+r_{{2}}\cos \left( \psi \right)\\ &r_1\,\sin(\varphi)=r_2\,\sin(\psi) \end{align*}

from here you obtain

\begin{align*} &x(\varphi)=\cos \left( \varphi \right) r_{{1}}+r_{{2}}\sqrt {1-{\frac { \left( \sin \left( \varphi \right) \right) ^{2}{r_{{1}}}^{2}}{{r_ {{2}}}^{2}}}} \end{align*}

The kinetic Energy is:

\begin{align*} &T=\frac{m}{2}\left(\frac{\partial x(\varphi)}{\partial\varphi}\,\dot{\varphi}\right)^2+\frac{M\,r_1^2}{2}\, \dot{\varphi}^2:=\frac{1}{2}\,g(\varphi)\dot{\varphi}^2 \end{align*}

and with Euler- Lagrange you obtain the equation of motion \begin{align*} &\ddot \varphi +\frac 12\,{\frac { \left( {\frac {d}{d\varphi }}g \left( \varphi \right) \right) {\dot\varphi }^{2}}{g \left( \varphi \right) } }=0 \end{align*} To obtain the constraint force of the rod $~r_1$, you "open a gap" $~q_n~$ towards $~r_1$ thus Kinematic \begin{align*} &x=\cos \left( \varphi \right) (r_{{1}}+q_n)+r_{{2}}\cos \left( \psi \right)\\ &(r_1+q_n)\,\sin(\varphi)=r_2\,\sin(\psi) \end{align*} From here you obtain $~x(\varphi~,q_n)$ you have now two generalized coordinates $~\varphi~,q_n$ The kinetic energy is

\begin{align*} &T=\frac{m}{2}\,v^2+\frac{M\,r_1^2}{2}\,\dot{\varphi}^2\\ &\text{with}\\ &v=\frac{\partial x}{\partial \varphi}\,\dot{\varphi}+ \frac{\partial x}{\partial q_n}\,\dot{q}_n\\\\ \Rightarrow\\ &T:=\frac{1}{2}\,g(\varphi~,q_n)\dot{\varphi}^2+ h(\varphi~,q_n)\,\dot{\varphi}\,\dot{q}_n+\frac{1}{2}\,f(\varphi~,q_n)\, \dot{q}_n^2 \end{align*}

you have to use now Euler- Language with holonomic constraint equation (you "close the gap") which is $q_n=0$ . you obtain the rod constraint force $~\lambda~$ \begin{align*} &\lambda=\left(\frac{1}{2}\frac{h(\varphi~,0)\,D_1(g)(\varphi,0)}{g(\varphi,0)}- \,D_1(h)(\varphi,0)+\frac 12 \frac{\partial}{\partial q_n}\,g(\varphi,q_n)\bigg|_{q_n=0}\right)\,\dot{\varphi}^2\\\ &\text{with}\\ &D_1(g)(\varphi,0)=\frac{\partial g}{\partial\varphi}\bigg|_{(\varphi,0)}\qquad ,D_1(h)(\varphi,0)=\frac{\partial h}{\partial\varphi}\bigg|_{(\varphi,0)}\\ \end{align*}


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