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## Homework Statement

A rod of mass M and length L is placed horizontally on a table such that the center of mass is a distance L/6 from the edge of the table (CoM is hanging off the table). The rod is released from rest and slips after reaching an angle θ. Find the coefficient of friction [itex]\mu[/itex]

**2. The attempt at a solution**

Friction is the only horizontal external force acting, so once the horizontal acceleration of the center of mass exceeds [itex]\frac{\mu F_N}{m}[/itex] the rod should slip.

Call the horizontal and vertical accelerations of the CoM [itex]a_x[/itex] and [itex]a_y[/itex] respectively. The normal force can be found from the relationship [itex]mg-F_n=ma_y[/itex] therefore the key constraint for the problem is:

[itex]\mu(mg-ma_y)=ma_x[/itex]

or

[itex]\mu(g-a_y)=a_x[/itex]

...

[itex]a_x=\frac{L}{6}ω^2\cos(\theta)+\frac{L}{6}α\sin(\theta)[/itex]

[itex]a_y=\frac{L}{6}α\cos(\theta)-\frac{L}{6}ω^2\sin(\theta)[/itex]

...

[itex]α=mg\frac{L}{6}\cos(\theta)[/itex]

[itex]ω^2=\frac{2}{I}mg\frac{L}{6}\sin(\theta)[/itex]

...

[itex]I=\frac{mL^2}{9}[/itex]

Sorry I don't have time at the moment to show all the intermediate algebraic steps or to explain each equation, but putting it all together I get

[itex]\mu=\frac{\cos(\theta)\sin(\theta)}{2-\cos^2(\theta)}[/itex]

Whereas the correct answer is supposed to be proportional to [itex]\tan(\theta)[/itex]

Are any of my equations wrong?