1. ## Rotational Dynamics

One-fourth length of a uniform rod of mass m and length l is placed on a rough horizontal surface and it is held stationary in horizontal position by means of a light thread. The thread is then burnt and the rod starts rotating about the edge. Find the angle between the rod and the horizontal when it is about to slide on the edge. The coefficient of friction between the rod and the surface is $\mu$.

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2. Done!

Using Parallel-axis theorem, Moment of inertia of the rod about pivot point:
$I_{P} = \frac{ml^2}{12} + m\left(\frac{l}{4}\right)^2 = \frac{7ml^2}{48}$
( $I_{CM} = \frac{ml^2}{12}$)

Imagine the rod is tilted at an angle $\theta$ w.r.t the horizontal. If you draw an F.B.D, there are three forces on the rod: $mg$(downward and acting on CM), $F_{\perp}$(reaction force - $\perp$ to the rod and acting on the pivot point outward), and $F_{fric}$( $\parallel$ to the rod and pointing to the left, but also upward at an angle $\theta$).

Considering the torque due to gravity about pivot point and hence finding angular acceleration as a function of $\theta$,

$\frac{1}{4}mgl\cos \theta = I_{P}\alpha(\theta)\ \Rightarrow \alpha(\theta) = \frac{12g}{7l}\cos \theta$

Considering torque due to reaction force about CM,

$F_{\perp}\cdot \frac{l}{4} = I_{CM}\alpha(\theta)\\ \Rightarrow F_{\perp} = \frac{4}{7}mg\cos \theta$

$F_{fric} = \mu F_{\perp} = \frac{4}{7}\mu mg\cos \theta$

Using conservation of energy principle and hence finding angular velocity as a function of $\theta$,

$mg\frac{l}{4}\sin \theta = \frac{1}{2}I_{P}\omega(\theta)^2\\ \Rightarrow \omega(\theta)^2 = \frac{24g}{7l}\sin \theta$

Considering the centripetal force that $F_{fric}$ provides,

$F_{fric} - mg\sin \theta = m\omega(\theta)^2\frac{l}{4}$
$\Rightarrow \frac{4}{7}\mu mg\cos \theta - mg\sin \theta = \frac{6}{7}mg\sin \theta$
$\Rightarrow \boxed{\theta = \arctan\left(\frac{4\mu}{13}\right)}$