 #1
lorenz0
 140
 26
 Homework Statement:

A mass ##m## can slide without friction inside a tube rotated around one of its ends in a horizontal plane with constant angular velocity ##\omega##. A second identical mass is hung from the mass by means of an ideal rope passing through the center of rotation as shown in the figure. To determine:
a) the radial distance ##R_0## from the center of rotation at which the first mass must be in order for the system to be in equilibrium;
b) the type of balance, giving adequate reasons for the answer;
c) the opposite lateral constraint reaction from the pipe wall in position ##R_0##.
d) supposing that the first mass starts from stationary from the position ##R_0## found, the opposite lateral constraint reaction from the pipe wall when the mass reaches a distance of ##2R_0##.
 Relevant Equations:
 ##F_{centr}=m\omega^2 r, F_g=mg##
1) To be in equilibrium, it must be $$\begin{cases}F_{centr}T=0\\ Tmg=0\end{cases}\Rightarrow F_{centr}=T=mg\Rightarrow m\omega^2 R_0=mg\Rightarrow R_0=\frac{g}{\omega^2}$$
2) It is intuitive that this equilibrium is unstable but I don't know how to formally prove this.
3) In ##R_0## the mass ##m## is in equilibrium so ##N_{lat}=0##
4) I have tried by using angular impulse theorem and the workenergy theorem but I haven't got anywhere so far. The result should be ##N_{lat}=\sqrt{2}mg##.