25th_November_2008

Now, consider the CM of any rod.

rXI(ang impulse) =same.

rXI=ΔL
rXI=Iw₁-Iw₀
where w₁=final angular vel
and w₀=initial ang vel

now for both rods w₀=0 . Thus the final ang. velocities should be the same.

i cudnt come online since 11:00!!
well here's my solution

let the rods be in contact for a short time interval dt
and let the force acting on both the rods(BY newtons third law) be F
then from the figure
we can see that amgular impulse imparted = FdtL/2 = ML²/12w1
again FdtL/2 = ML²/12w2

so w1 =w2
further due to Fdt

for rod 1

Fdt = mv2 (v2 final velocity for rod in rest intitially)

for rod2

Fdt = m(v-v1) (v1 final velocity of rod 2)

and further it is given that coefficient of restitution = e

now here the problem arises, e can be equated to (rel. velocity after collision/rel. velocity before collision of the bodies)

but , does the term bodies mean the centre of mass or the point of impact?
most probably it is the point of impact as it takes most actively part int the collision
due to e it gets deformed thus in my view bodies = > point of impact

here , rel. velocity of point of impact before collision =
v
after collision =>

v1+wL/2 + wL/2-v2

so ev= v1-v2+wL/2

and further
mv=mv1+mv2 => v=v1+v2
we get =

v1=(v+ev-wL)/2

but we know
Fdt=m(v-v1)
and FdtL/2 = M(L²/12)w

so we get the value of w and v1 and v2
subsequently

okie.............the left rod will rotate in counterclockwise dirn along a axis thru its perpandicular bisector and the right rod will rotate in counterclockwise sense but bout the axis thru its left end and both will have same value of w

Firstly........ both will have same ω as.........

The situation has an element of symmetry....
It will not make a difference whether this rod is coming or other one...