 #1
 81
 2
Poster has been reminded to use the Template when starting schoolwork threads
The coordinates of the mass 'm' is $$x=R\cos(\omega{t})+R\cos(\omega{t}+\phi)$$ $$y=R\sin(\omega{t})+R\sin(\omega{t}+\phi)$$
Where $\phi$ is the angle of the particle with respect to coordinate system attached to the circle with origin at center of the circle.
Since particle is not acted upon by any force, So the potential energy can be taken as zero.
The Lagrangian of the system is
$$L=\frac{m}{2}\bigg((R\omega)^2+R^2(\omega+\dot{\phi})^2+2R\omega(\omega+\dot{\phi})\cos{\phi}\bigg)$$
$$J=\frac{\partial L}{\partial \dot{\phi}}=mR^2(\omega+\dot{\phi})+mR^2\omega\cos{\phi}$$
So the equation of motion for $\phi$ coordinate is
$$\ddot{\phi}+{\omega}^2\sin\phi=0$$.
This is the result. I have no doubt regarding this problem. I have doubt regarding total energy of this problem, since the particle is not subjected to any potential, we took $U$=0. So Lagrangian is simply Kinetic energy of the particle, The total energy is the sum of Kinetic energy and potential energy(U=0 for this case), can I infer the Total energy and Lagrangian has the same form?
My actual doubt is Total energy of the particle is not constant of motion, rather $E\vec{\omega}.\vec{J}$ is the constant of motion. when I derived I got
$$C=E\vec{\omega}.\vec{J}=\frac{m}{2}\bigg(R^2\dot{\phi}^2+R^2\omega\dot{\phi}\cos{\phi}\bigg)$$
By the definition of Constant of motion, if I take the time derivative of the above quantity, it should give me zero. But I am getting
$$\dot{C}=mR^2\bigg((\dot{\phi}\omega^2+\omega\dot{\phi}^2)\sin\phi+\omega^3\sin\phi\cos\phi\bigg)\neq{0}$$
I used $$\ddot{\phi}= {\omega}^2\sin\phi$$ to simplify the above expression.
Is there anywhere I did a mistake, since potential energy is taken as zero, I took total energy as same as Lagrangian, is it wrong? or I messed up somewhere?
Attachments

9.4 KB Views: 432
Last edited by a moderator: