 #1
LCSphysicist
 634
 153
 Homework Statement:

A uniform cylinder of radius a and density ρ is mounted so as to rotate freely
about its axis, which is vertical. On the outside of the cylinder is a rigidly fixed uniform spiral
or helical track along which a point mass m can slide without friction. Suppose the point mass
starts at rest at the top of the cylinder and slides down under the influence of gravity.
FInd the lagrangian of the (cylinder+bead) system
 Relevant Equations:
 .
Hello. I am having some difficulties regarding this problem.
I am not sure, actually, how to relate the angle rotated by the cylinder ##\phi## and the angle drawed by the bead itself ##\theta##.
I have been thinking in two alternatives:
a) The first one is that, since there are no friction forces, the angle are independent. So the Kinect energy of the whole system would be $$\frac{I \dot \phi ^2}{2} + \frac{m(\dot z ^2 + (r (\dot \theta + \dot \phi) ) ^2)}{2}$$
b) The second alternative would be to consider that, since gravity itself acts only at the z direction, and the rest of forces are internal. ##L_{z (cylinder)} = L_{z( bead)}##
I am not sure which one is correct, if neither.
I am not sure, actually, how to relate the angle rotated by the cylinder ##\phi## and the angle drawed by the bead itself ##\theta##.
I have been thinking in two alternatives:
a) The first one is that, since there are no friction forces, the angle are independent. So the Kinect energy of the whole system would be $$\frac{I \dot \phi ^2}{2} + \frac{m(\dot z ^2 + (r (\dot \theta + \dot \phi) ) ^2)}{2}$$
b) The second alternative would be to consider that, since gravity itself acts only at the z direction, and the rest of forces are internal. ##L_{z (cylinder)} = L_{z( bead)}##
I am not sure which one is correct, if neither.