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## Homework Statement

Rigid uniform Rod 'AB' of mass 'M' and Length 'L' is pulled slightly ( Gently ) at the bottom at time t=0 when it just reaches the horizontal ground, at that time find the Velocities of End Points of the rod.

All the surfaces are friction less.

## Homework Equations

## The Attempt at a Solution

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I first tried to find the value of ##\theta## at which the rod left contact with the vertical wall.

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}}##

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

After loosing contact with the vertical wall no horizontal force acts on the CoM of the rod so the vertical component of the rod will not change.

So ##V_{A}=\frac{\sqrt{gl}}{3}##. Where ##V_{A}## is the velocity of the lower end of the rod when the rod becomes horizontal.

Let the rod make an angle ##\phi## withe the horizontal after loosing contact.

So ##-\frac { Nlcos\theta }{ 2 } =\tau ##

or ##\frac { Nlcos\phi }{ 2 } =-I\frac { d\omega }{ d\phi } \omega ##

or ##\frac { Nlsin\phi }{ 2 } =-I\frac { { \omega }^{ 2 } }{ 2 } +C##

At ##sin\theta=2/3 {\omega}^{2}=\frac{g}{l}##

So ##C=\frac { l(8N+mg) }{ 24 } ##

So ##\frac { Nlsin\phi }{ 2 } =-I\frac { { \omega }^{ 2 } }{ 2 } +\frac { l(8N+mg) }{ 24 }##

When the rod becomes horizontal ##\theta=0##

So ##N=\frac { m(l{ \omega }^{ 2 }-g) }{ 8 } ##

Now the acceleration of the CoM of the rod(##a##) is ##\frac{mg-N}{m}##

vertical velocity of CoM when rod left contact with the wall is ##\frac{\sqrt{5gl}}{2}##

distance covered by the CoM =##\frac{l}{3}##

So vertical velocity of the CoM of the rod when it becomes horizontal(##v##) is

## v^{2}=\frac { 5gl }{ 36 } +\frac { 2l(mg-N) }{ 3m } ##

Also when the rod becomes horizontal ##\frac{l\omega}{2}=v##. As the vertical velocity of point A is 0.

So on putting values of ##N## and ##\omega##

I got ##4v^{2}=\frac{13gl}{6}##

Also the ##V_{B}=2v##

So ##V_{B}=\sqrt{\frac{13gl}{6}}##.

Is my approach correct? Are my answers correct?

Please help.:)

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