I tried to recap everything and I think I spotted some mistakes.. could you please read it and tell me if it sounds good,
@TSny ? Please!
##d_{cm}=\frac{\frac{L}{2}\cdot2M +0 \cdot M}{3M}=\frac{L}{3} ##
##v_{cm}= \omega d_{cm}##
At start the horizontal linear momentum is ##3Mv_{cm_{0}}##, while at ##\theta_{max}## rod-and-sleeve is just translating at some velocity V, so the hor. linear momentum is ##3MV##.
As the h. linear momentum conserves itself (there are no horizontal forces), it must be ##V=v_{cm_{0}} = \omega_{0}\frac{L}{3}##.
There are no dissipative forces, so U + K conserves.
At first (##\theta=0##), instead the motion is entirely rotational: ##E_{\theta=0}=\frac{1}{2}I\omega_0^2##, where I is the moment of inertia, calculated as ##I=\frac{1}{3}(2M)L^2##.
At ##\theta_{max}##, the motion is entirely translational, so ##E_{\theta_{max}}## is ##\frac{1}{2}(3M) V^2+ L(2M)g(1-\cos(\theta_{max}))## (where the second term is the expression for the potential energy.
From ##E_{\theta_{max}}=E_{\theta=0}## I get ##\cos(\theta_{max})=1-\frac{5L\omega_0^2}{12g}##.
Then, using the following notation (measured in the zero-momentum frame):
U is the instantaneous velocity of the sleeve. Positive toward the right.
vx is the x-component of the center of mass of the rod. Positive x is to the right.
vy is the y-component of the center of mass of the rod. Positive y is upward.
θ is the angular displacement of rod from horizontal. Counterclockwise is positive.
R is the distance pivot-center of mass (##d_{cm}##)
ω is the angular velocity of the rod. Counterclockwise is positive.
vx=U+ωRcosθ
In the zero-momentum frame MU+2Mvx is constantly 0.
M(U)+2M(U+ωRcosθ)=0 hence ##U = -2/3 ω R \cos(θ)##.
While the rod is swinging up it loses horizontal momentum, so also the sleeve is reducing its horizontal momentum (talking about abs values). While the rod swings down, instead, it gains horizontal momentum, and so does the sleeve. What we want to verify is that effectively the sleeve has its maximum KE at ##\theta=0## (we want to show that |U| never decreases as |θ| decreases).
Let's recap:
vx = -U/2 (from conservation of momentum)
##U = -\frac{2}{3}ωRcosθ##
vy = ωRsinθ
If |U| were to decrease as |θ| decreased, |vx| decreases, |ω| decreases, |vy| decreases.. that means that if |U| decreases when |θ| decreases, then KE of the system decreases, but this is not possible because also potential energy is decreasing and there are no dissipative forces. OK, we finally found out that max sleeve speed is when the rod comes back in the vertical position.
If θ max, with vx the velocity in the original frame:
- in the zero momentum frame the horizontal momentum is 3M*(vx - w) = 0 i.e. w=vx (w is the transformation factor between the frames).
I conclude that w is V.
##U_{max} = -2/3 ω_{θ=0} R cosθ + w##
Due to the symmetry of motion in the zero-momentum frame ##ω_{θ=0}## is opposite to ω0 in every inertial frame and
##U_{max} = -2/3 ω_{θ=0} R cosθ + w = -2/3 ω_{0} R + ω_0 R=1/3 ω_0 R##
We'll either write L/2 or L/3 explicitly.
), even if I have seen that it just works 
