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Consider a thin uniform rod AB of mass $m$ and length $l$ sliding down a frictionless wall and on a frictionless floor. Point O is its instantaneous center of rotation, and C its center of mass (and geometric center). We need to calculate the angle $\theta$ after which the rod loses contact with the wall. The initial value of $\theta$ is given to be $\pi/3$ radians.

I've marked the directions of motion of the end points and the center of mass of the rod (pardon my MS paint skills). The forces on the rod are its weight through C, $N_{wall}$ is the reaction force from the wall and $N_{ground}$ is the reaction force from the wall.

The problem for me is that calculating the angular acceleration $\alpha$ about O gives me

$ (ml^2/3) \alpha = mgl(cos{\theta})/2$

And when we do the same from A, we get

$ (ml^2/3) \alpha = mgl (cos{\theta})/2 - N_{wall}lcos{\theta}$

And from B, we get

$ (ml^2/3) \alpha = N_{ground}lcos{\theta} - mgl (cos{\theta})/2 $

I'm having trouble figuring out where I'm going wrong. Supposedly, the angular acceleration for a rigid body about all points should be the same. Which of the above equations is wrong, then? Is it because the points in question are non-inertial/accelerating?

As far as the question itself is concerned, I think I can do it based on making a differential equation based on the given equation(s) and horizontal acceleration being zero when it does lose contact. I just think that there's some fundamentally wrong with the way I've approached the problem above, and would really appreciate some help. Thanks!

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closed as off-topic by John Rennie, Emilio Pisanty, knzhou, rob Apr 23 '18 at 11:50

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  • $\begingroup$ Is there friction? $\endgroup$ – Tom B. Apr 22 '18 at 19:38
  • $\begingroup$ no, edited that now $\endgroup$ – Attila1177298 Apr 22 '18 at 19:40