0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $$z_{max}-z_{0}=\frac{mv_{0}^{2}(1+\cot^{2}{\alpha})\tan^{2}{\alpha}}{2g}$$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.