Angular velocity question?

Question

A thin rod has a length of 0.831 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.894 rad/s and a moment of inertia of 1.14 x 10-3 kg·m2. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (whose mass is 5 x 10-3 kg) gets where it's going, what is the change in the angular velocity of the rod?

Answer 1

For a rod Ir= (m1R^2)/6
For the insect Ii=m2R^2

Angular momentum
L(rod)=Ir w

L(insect)= R mV (cross product implied)
V=wR
L(incect)=R m w R= m wR^2

Momentum conservation equition
Ir w' =Irw + m2wR^2 where
w' angular speed =0.894 rad/s
w -?

[(m1R^2)/6]w'=[(m1R^2)/6]w +m2wR^2
m1w' = m1w+ 6m2w

Now we have
w=m1w'/(m1 +6m2)
m1=6 Ir/R^2 =6 x 1.14 E-3/(0.831 /2)=0.0165 kg

w=0.0165 x 0.894 /(0.0165 + 6 x 0.005 )
w=0.317 rad/s

change in angular speed is
dw=0.831 -0.317=0.514 rad/s