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powerof

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## Homework Statement

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As a diagram any of the countless images on the internet serve well:

Suppose the static friction was not sufficient to maintain the rod in a static equilibrium and it starts slipping/sliding. It has an initial angle theta with respect to the ground. Assume there is no dynamic friction and that the rod has uniform mass. The rod has a length L. With what velocity will it's center of mass hit the ground?

## Homework Equations

The moment of inertia is useful in this case: [itex]I^*=\frac{1}{12}MR^2[/itex]

[itex]\tau_{total} = I^* \alpha [/itex]

## The Attempt at a Solution

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I tried analyzing the forces and then looking at the torque as well.

Y axis (positive up).

[itex]N_{ground}-Mg=Ma_{CM_{y}} \Rightarrow N_{ground} = M(g+a_{CM_{y}})[/itex]

X axis (positive right).

[itex]N_{wall}=Ma_{CM_{x}}[/itex]

Torques with respect to the center of mass (positive counterclockwise, or "out of the screen"):

[itex]\tau_{total} = I^* \alpha = N_{ground}\frac{L}{2}cos\theta - N_{wall}\frac{L}{2}sin\theta = \frac{1}{12}ML^2\alpha[/itex]

[itex] M(g+a_{CM_{y}})\frac{L}{2}cos\theta - Ma_{CM_{x}}\frac{L}{2}sin\theta = \frac{1}{12}ML^2\alpha[/itex] //substituted the normal forces

[itex] (g+a_{CM_{y}})cos\theta - a_{CM_{x}}sin\theta = \frac{1}{3}\frac{L}{2}\alpha = \frac{1}{3} (a)_{t}[/itex] //a_{t} is the acceleration of a point at the end of the rod in the same direction as the same point's velocity

I'm more or less stuck here. How can I relate the velocity of the center of mass with something I know?

Thank you for your time.

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