Recent content by Physics_Enjay

Thanks for your help guys. H(z0)=H(z1), where z1 and z0 are extrema of z (ie z' and hence p_z are zero), and solve for z1 in terms of z0. I found the question confusing because I misread it as though z in the question was a general z.

Isn't it because Ice skater changes his or her moment of inertia? In this problem we have a point particle.

This is my reasoning: We said that angular momentum is constant, which implies that angular velocity is constant, which implies that angular acceleration ##\phi''(t)## is zero, i.e. ##\phi''(t) = \frac{p_{\phi }'(t)}{m z(t)^2}-\frac{2 p_{\phi }(t) z'(t)}{m z(t)^3} = 0##. The first term is zero...

Sorry I forgot to include the minus for ##p_z '(t) ## so ##p_z '(t) = - (g m-\frac{p_{\phi }^2}{m z(t)^3})## and hence the previous result is ##z(t) = (\frac{p_{\phi }^2}{gm^2})^{1/3}##. I'm still not sure how to get further with this. I already used the fact that the Hamiltonian is conserved...

##\phi''(t) = \frac{p_{\phi }'(t)}{m z(t)^2}-\frac{2 p_{\phi }(t) z'(t)}{m z(t)^3} = 0## implies that z'(t) =0 always, not just at extrema, right? This then implies that ##p_z (t) = 0## which gets rid of that term in the Hamiltonian. Also have that ##p_z '(t) = g m-\frac{p_{\phi }^2}{m...

Sorry, what I meant to say was that because angular velocity is constant, then ##\phi '(t)## is constant

Since energy is conserved in the system the particle doesn't come to rest. If the particle moves between those values for z, then they are extremas of z, so I know that ##z'(t)## is zero. I'm not sure what this says for ##\phi## but I know that since angular momentum is conserved then so is...

Homework Statement A particle slide on the frictionless surface of the interior of a 45 degree cone ##x^2 + y^2 = z^2 ## a) Find the 2D Lagrangian in terms of the vertical coordinate ##z## and an angular coordinate ## \theta ##. b) Find the Hamiltonian ##H##. c) Show that ##p_\theta## and...