A rod released on a rough floor

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Taking torque about point of contact

$\frac { m{ l }^{ 2 } }{ 3 } \alpha =\frac { l }{ 2 } mgsin\theta$

But $a=\frac{l \alpha}{2}$

So $a=\frac{3g sin \theta}{4}$

Let the velocity of CoM of rod be $v$ at the moment rod makes angle $\theta$ with the vertical. So $\omega=\frac{2v}{l}$

Also by conservation of energy

$mg\frac { l }{ 2 } (1-cos\theta )=\frac { 1 }{ 2 } I{ \omega }^{ 2 }$

On putting the value of $\omega$ we will get

$\frac { 3g }{ 2 } (1-cos\theta )=\frac { 2{ v }^{ 2 } }{ l }$

Also $\frac { 2{ v }^{ 2 } }{ l } =a_{c}$

From figure 2

$f=m(a cos \theta-a_{c} sin \theta)$

But $f=0$ so on putting values

$\frac{3g sin \theta cos \theta}{4}=\frac{ 3g (1-cos\theta )sin \theta}{2}$

On solving we will get

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

So $\frac { 2 }{ 3 } \times 24=16$

Please re-share this question guys. I think it is underrated.