When two masses strike the rod, then angular momentum imparted to rod,
L1+L2=2mv(L)+m(2v)(2L) =6mvL Now, after striking of masses to rod the angular momentum of complete rod about the centre
C ,
Lrod =10 where,
ω= angular velocity of the rod and
I= moment of inertia of the rod.
The moment of inertia of rod
I===24mL2 ∴M=8m and
I=6L (given)
Now, moment of inertia of two masses after the striking to \/ rod
l1=2m(L)2=2mL2 and
I2=m(2⟂)2=4mL2 ∴ The net moment of the inertia about the centre of rod,
I=24mL2+2mL2+4mL2 ⇒=30mL2 By putting this value in Eq. (i), we get
Lrod =30ML2ω From the law of conservation of angular momentum,
L1+L2=Lod ⇒6mvL=30mL2ω ⇒ω= Hence, the anguler velocity of the rod is
.