Will Sliding Rod ever Stop ? (Part 2)

First we have to find the angle at which the rod left contact with the vertical wall/

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

So at $\theta=arcsin(2/3)$ , $\omega=\sqrt{\frac{g}{L}}$

and ${ V }_{ C }=\frac{\sqrt{gl}}{3} { i } -\frac{\sqrt{5gl}}{6} { j }$

Now

When the rod becomes horizontal point A will not have any vertical velocity. So vertical component of velocity of CoM is equal to angular velocity of point A about CoM. So being in opposite direction they cancel out. So remaining vector is the horizontal velocity of CoM. So velocity of pointA equal to the component of the horizontal velocity of the CoM.

So $V_{A}=\sqrt { \frac { gl }{ 9 } }$

Now further using conservation of energy we get

The initial velocity of the rod is $\frac { l\omega cos{ (\phi }_{ 0 }) }{ 2 } =\frac { \sqrt { 5gl } }{ 6 }$ where $\phi_{0}$ is the angle b/w rod and horizontal at the moment the rod looses contact with the vertical wall.

Initial potential of the rod is $\frac{lsin\phi_{0}}{2}=\frac{l}{3}$

Let $v'$ be the velocity of the CoM of the rod when it becomes horizontal.

So by the conservation of mechanical energy---

So $\frac { { I }_{ c }{ \omega }_{ 0 }^{ 2 } }{ 2 } +\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } +\frac { { I }_{ c }{ \omega }_{ f }^{ 2 } }{ 2 }$

where $\omega_{ 0 }$ is the angular velocity of the rod when it looses contact with the vertical wall and $\omega_{ f }$ is the angular velocity of the rod when it become horizontal.

or $\frac { m{ l }^{ 2 } }{ 12 } \times \frac { { \omega }_{ 0 }^{ 2 } }{ 2 } +\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } +\frac { m{ l }^{ 2 } }{ 12 } \times \frac { { \omega }_{ f }^{ 2 } }{ 2 }$

or $\frac { 32gl }{ 36 } ={ v' }^{ 2 }+\frac { { l }^{ 2 }{ \omega }_{ f }^{ 2 } }{ 12 }$ .

When the rod becomes horizontal then the vertical component of velocity of point A(the point which was in contact with the ground) will be zero.

So is $\frac { l{ \omega }_{ f }^{ 2 } }{ 2 } ={ v' }$

So ${ \omega }_{ f }=\sqrt { \frac { 8g }{ 3l } }$

and $v'=\frac { l }{ 2 } \sqrt { \frac { 8g }{ 3l } }$

velocity of point B is equal to total velocity of CoM+velocity of point B about CoM

So total velocity is

${ V }_{ CoM }=-\frac { l }{ 2 } \sqrt { \frac { 8g }{ 3l } } \hat { j } +\sqrt { \frac { gl }{ 9 } } \hat { i }$ $+\left( -\frac { l }{ 2 } { \omega }_{ f }\hat { j } \right)$

or $\sqrt { \frac { gl }{ 9 } } \hat { i } -\sqrt { \frac { 8gl }{ 3 } } \hat { j }$

it's magnitude will be $\sqrt { \frac { 25gl }{ 9 } }$

So $V_{B}=\sqrt { \frac { 25gl }{ 9 } }$ . :)