A particle of mass m moves without friction on a plane making an angle of alpha with the horizontal (earths surface).

How do I determine Lagrange's equations of motion for this system?

Given that the particle starts at the origin with a velocity $v(0) = v_0(\hat{i} + \tan \alpha \hat{k})$, with $v_0 > 0$. Determine the maximum height reached by the particle and the time taken for the particle to return to earth.

  • $\begingroup$ Just find the lagrangian $L=T-V$ first by placed the coordinate $x-z$ so that the vertex of the inclined plane coincide with origin. $\endgroup$ – Sou Sep 24 '17 at 3:37

The Lagrangian of a particle in gravitational acceleration is $$\mathcal{L}=\frac{m}{2}(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2})-gz$$ The equation of a plane inclined by $\alpha$ may be written as $$x\sin(\alpha)-z\cos{\alpha}=0$$ Thus $$\dot{x}=\dot{z}\cot{\alpha}$$ So $$\mathcal{L}=\frac{m}{2}(\dot{y}^{2}+[1+\cot^{2}{\alpha}]\dot{z}^{2})-gz$$ $y$ is a cyclic variable, so $$p_{y}=m\dot{y}=const$$ Thus $$\mathcal{L}=\frac{m}{2}\Big(\frac{p_{y}^{2}}{m^{2}}+[1+\cot^{2}{\alpha}]\dot{z}^{2}\Big)-gz$$ $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial{\dot{z}}}=\frac{\partial\mathcal{L}}{\partial{z}}$$ $$\ddot{z}+\frac{g}{m[1+\cot^{2}{\alpha}]}=0$$ $$z(t)=-\frac{g}{2m[1+\cot^{2}{\alpha}]}t^{2}+v_{z0}t+z_{0}$$ $$z(t)=-\frac{g}{2m[1+\cot^{2}{\alpha}]}t^{2}+v_{0}\tan{\alpha}t+z_{0}$$ To find the maximum distance, extrimize $$\frac{d}{dt}z(t)=0$$ $$t=\frac{m[1+\cot^{2}{\alpha}]v_{0}\tan{\alpha}}{g}$$ Hence, the maximum height is $$z_{max}-z_{0}=\frac{mv_{0}^{2}(1+\cot^{2}{\alpha})\tan^{2}{\alpha}}{2g}$$$


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