Sliding Rod!

we can solve this question by IAOR ( Instantaneous Centre of rotation )

Using energy conservation in ICR frame : ( Since Rod is in pure Rotational motion about ICR )

$\cfrac { mgL }{ 2 } (1-\sin { \varphi } )\quad =\quad \cfrac { 1 }{ 2 } { I }_{ ICR }{ \omega }^{ 2 }\\ \cfrac { mgL }{ 2 } (1-\sin { \varphi } )\quad =\quad \cfrac { 1 }{ 2 } (\cfrac { m{ L }^{ 2 } }{ 12 } +\cfrac { m{ L }^{ 2 } }{ 4 } ){ \omega }^{ 2 }\\ { \omega }^{ 2 }=\quad \cfrac { 3g }{ L } (1-\sin { \varphi } )\quad \quad \quad .\quad .\quad .\quad (1)$ .

$\alpha =\quad \omega \cfrac { d\omega }{ d\varphi } \quad =\quad -\cfrac { 3g }{ 2L } (\cos { \varphi } )\quad \quad .\quad .\quad .\quad .\quad (2)\\ { a }_{ C }=\quad { \omega }^{ 2 }\cfrac { L }{ 2 } \quad =\quad \cfrac { 3g }{ 2 } (1-\sin { \varphi } )\quad \quad .\quad .\quad .\quad .\quad (3)\\ { a }_{ T }=\quad \cfrac { L }{ 2 } \alpha \quad =\quad -\cfrac { 3g }{ 4 } (\cos { \varphi } )\quad \quad \quad .\quad .\quad .\quad .\quad .(4)$ .

Rod will Loose it's Contact with the wall if ${ N }_{ 1 }\quad =\quad 0\\ \\ \Rightarrow \quad { a }_{ com,net,\quad Horizontal }\quad =\quad 0$ .

Resolve net acceleration in vertical and horizontal direction and equating horizontal component to zero , so we should get ,

$\sin { \theta } \quad =\quad \cfrac { 2 }{ 3 } \quad ,\quad \\ \Rightarrow \boxed { \tan { \theta } \quad =\quad \cfrac { 2\sqrt { 5 } }{ 5 } }$ . Q.E.D

This is second time I am posting this figure. :D

img From the figure

${ V }_{ BC }=\frac { l\omega }{ 2 }$

${ V }_{ BC }=\frac { l\omega }{ 2 } sin\theta \hat { i } +\frac { l\omega }{ 2 } cos\theta \hat { j }$

and ${ V }_{ AC }=-\frac { l\omega }{ 2 } sin\theta \hat { i } -\frac { l\omega }{ 2 } cos\theta \hat { j }$

As ${ V }_{ BC }={ V }_{ B }-{ V }_{ C }$ and ${ V }_{ AC }={ V }_{ A }-{ V }_{ C }$

or ${ V }_{ C }={ V }_{ B }-{ V }_{ BC }$

So ${ V }_{ C }\hat { j } =-\frac { l\omega }{ 2 } cos\theta \hat { j }$ (considering velocity in vertical direction only. We are doing this because ${ V }_{ B }\hat { j } =0$ )

Similarly we can get

${ V }_{ C }\hat { i } =\frac { l\omega }{ 2 } sin\theta \hat { i }$ ............(4)

So ${ V }_{ C }=\frac { l\omega }{ 2 } sin\theta \hat { i } -\frac { l\omega }{ 2 } cos\theta \hat { j }$

Also from conservation of energy

$mg\frac { l }{ 2 } (1-sin\theta )=\frac { 1 }{ 2 } \left( m{ v }^{ 2 }+{ I }_{ C }{ \omega }^{ 2 } \right)$ ......................(1)

Also ${ V }_{ C }={ \left( \frac { l\omega }{ 2 } \right) }^{ 2 }$ (we can get this from eq(4))

Putting this in eq-1 we get

${ \omega }^{ 2 }=\frac { 3g }{ l } (1-sin\theta )$ ............(2)

When the rod looses contact with the vertical rod the Normal force acting on Rod becomes 0. So no horizontal force act on the rod after the moment the rod left contact with vertical wall. So acceleration of the CoM of the rod in horizontal direction is 0.

So ${ \frac { d }{ dt } (V }_{ C }\hat { i } )=\frac { d }{ dt } \left( \frac { l\omega }{ 2 } sin\theta \hat { i } \right) =0$

So $\frac { d }{ d\theta } \left( \sqrt { \frac { 3g }{ l } (1-sin\theta ) } sin\theta \right) \frac { d\theta }{ dt } =0\\$

or $\frac { d }{ d\theta } \left( \sqrt { (1-sin\theta ) } sin\theta \right) =0$

or $cos\theta \sqrt { (1-sin\theta ) } =\frac { sin\theta cos\theta }{ 2\sqrt { (1-sin\theta ) } }$

So $sin\theta =\frac { 2 }{ 3 }$

or $tan\theta =\frac { 2\sqrt { 5 } }{ 5 }$ . :)