doubt mechanicss

is the answer

2v₀/[l cos(theta)]

firstly write torque equation about point of rotation i.e

Mgl/2SIN@=1/3ML^2a[a is the angular accleration}

now replace a by dw/dt and multiply both sides by d@

3g/2l∫SIN@=∫wdw(CAN BE OBTAINED BY E CONSERVATION ALSO)

find w of rod just before collision.

now apply angular momentum conservation about that point.

it will have only w after collision since impact force is Fdt at centre of mass. this way you find w.

see initially i assumed both I_x and v_x

abhi agar I_x zero hota hai to v_x bhi apne aap automatically zero ho jaega,see equation 2

and after dt time , velocity is imparted by the impulse itself

and see, u cant just write the torque equation about that lowest point as Iα ,
i will tell u why

see angular momentum of a rigid system about any point P can be written as (angular momentum about P due to pure translational motion of centre of mass ) + (angular momentum due to pure rotational motion of COM abt P )

i.e
L_{p}=L_{cm,about P(trans)}+L_{cm,about P(rot)}
i.e
L_{p}=I_{cm}\omega +\vec{r}_{c/p}\times m\vec{v}_{c}

for this sum , pt P is the lowest pt of the rod so distance between COM and P remains same

differentiate both sides wrt time
\tau _{p}=I_{cm}\alpha +\vec{r}_{c/p}\times m\vec{a}_{c}

\tau _{p}=I_{cm}\alpha +\vec{r}_{c/p}\times m(\vec{a}_{c/p}+\vec{a}_{p})

\tau _{p}=I_{cm}\alpha +(\vec{r}_{c/p}\times m\vec{a}_{c/p}+\vec{r}_{c/p}\times m\vec{a}_{p})

now cross product of component of a_c/p parallel to r_c/p will be
zero, and component of ac/p perpendicular to r_c/p is simply r_c/pα

so \vec{r}_{c/p}\times m\vec{a}_{c/p} = mr^{2}_{c/p}\alpha

so

\tau _{p}=(I_{cm} +m{r}_{c/p}^{2})\alpha +\vec{r}_{c/p}\times m\vec{a}_{p}
\tau _{p}=I_{p}\alpha+\vec{r}_{c/p}\times m\vec{a}_{p}

here since pt P is accelerated u cant just simply write torque abt P = I_pα

experts n e views ?

let the impulse be J
and final velocity be v in the upward direction

\mathrm{J}=m(v_0+v)\cdots\ \ \boxed1\\ \frac{\mathrm{J}\cos\theta l}{2}=\frac{ml^2}{12}\omega\ \ \cdots\boxed2\\ \texttt{since the collision is elastic}\\ v+\frac{\omega l}{2} \cos\theta=v_0\\ \frac{J}{m}+\frac{3J\cos^2\theta}{m}-v_0=v_0\\ J=\frac{2mv_0}{1+3\cos^2\theta}\\ \texttt{from here we get omega by back substituting}\\ \omega=\frac{12v_0\cos\theta}{l(1+3\cos^2\theta)}

ok ok...