doubt mechanicss

is the answer

2v₀/[l cos(theta)]

You have found the w at the given instant.We need to find it after collision.

guyzz m not having the answer a frnd of mine gave it to me......

I_y = m (v_y + v_o) ........(1)

I_x=mv_x..........(2)

also
writing angular impulse abt com

I_yLcosθ2 - I_xLsinθ2 = mL²12w...........(3)

since collision is elastic , energy is conserved

mv_o²2= mv_c²2+ 12I_cw² = m(v_x²+v_y²)2 + 12I_cw² .........(4)

now since friction isnt present I_x= 0 ...........(5)

now unknowns are v_x ,v_y,I_x,I_y, w
i.e 5 unknowns and we hav 5 eqns

so u can find w

( i m not sure abt the 5th equation , experts pls confirm )

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

I didnt get u

i ve not seen the soln yet ..ll revise the chap again nd then ll try myslf nd then go through the soln btw thnxx qwerty bhaiiya nd all otherss....experts can also give their 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)}

cont from #6

v_o+v= Lw6cosθ =I_ym

v_o² - v²=L²w²12

(v_o - v) ( Lw6cosθ ) = L²w²12

v_o - v = Lwcosθ2

2v_o = Lw6cosθ+Lwcosθ2 =Lw (2+6cos²Î¸)12cosθ

w = 12v_ocosθL(1+3cos²Î¸)
i m getting slightly different

ok ok...