Rotational velocity of a system that has a mass launched at it stick to it

[ttk3]

Homework Statement

On a frictionless table, a glob of clay of mass 0.74 kg strikes a bar of mass 1.90 kg perpendicularly at a point 0.48 m from the center of the bar and sticks to it. If the bar is 1.50 m long and the clay is moving at 6.8 m/s before striking the bar, at what angular speed does the bar/clay system rotate about its center of mass after the impact?

Homework Equations

displacement of center of mass = m1x1 + m2x2 \ m1 + m2
Ip = Icm + md^2
mVo = IW

The Attempt at a Solution

displacement of center of mass = 1.90*.48 / .74 *.190
= .6486

Ip = 1.90(.75^2) + (.74+1.90)(.48)
= 2.34

mcVo = IW
.74(6.8) = 2.34 W
= 2.15 rad/s

 

[saket]

Homework Statement

Homework Equations

displacement of center of mass = m1x1 + m2x2 \ m1 + m2
Ip = Icm + md^2
mVo = IW

I think u should check the last equation. Furthermore, please apply conservation of angular momentum carefully.