Finding angular speed of a system about its center of mass after the impact

[Smartguy94]

Homework Statement

1. Homework Statement
On a frictionless table, a glob of clay of mass 0.72 kg strikes a bar of mass 1.34 kg perpendicularly at a point 0.23 m from the center of the bar and sticks to it.

If the bar is 1.22 m long and the clay is moving at 8.3 m/s before striking the bar, what is the final speed of the center of mass?

At what angular speed does the bar/clay system rotate about its center of mass after the impact?

2. The attempt at a solution

Here is what I did

Vcm = sum(mi*vi) / total mass

in this case just one body has velocity:
Mclay = .72 kg
Mbr = 1.34 Kg
Mtotal = 2.06 kg
Vclay = 8.3 m/s

Vcm = 2.901 m/s

which is correct, I got the first part, but here is the second part that I got wrong

now calculate the angular speed:

angular speed - omega

omega = v /r

v - is the linear velocity on the trajectory of the body (tangential velocity)
r - is the distance between the body which rotate and the center of rotation

in your case the centre of rotation is the center of mass and r is the distance of the clay to the centre of mass

D = 1.22 m
d = 0.23 m

the centre of the bar related to one end is D/2
the position of the clay related to the same end is d+D/2

Xcm = [Mclay *(d+D/2)+Mbr*(D/2) ]/Mtotal

Xcm = .6904 m position of the center of mass

the angular velocit of the clay:

omega clay = Vclay / D1

D1=(d+D/2)-Xcm = .1496 m

omega clay = 55.477 rad/s

now about the bar:

the center of the bar is situated related to the center of mass at:

Dbar = Xcm - D/2 = .0804 m

If you assume that the bar when it rotate has the same tangential velocity v = 8.3 m/s

omega bar = 8.3 / 0.0804 = 103.249 rad/s

the question is asking about At what angular speed does the bar/clay system rotate about its center of mass after the impact?

and so I add up both the omega of the bar and the clay and got 158.726 rad/s
but it's wrong

Can anyone tell me where my mistake is?

 

[Simon Bridge]

You cannot have the bar and the clay with different omega.
You got that because you assumed you had the same tangential speed for the com bar and the clay - but surely the clay will be moving slower after impact?

This is a conservation of momentum problem.

The com speed formula you used is derived from the conservation of linear momentum - you then need to conserve angular momentum.

What is the total angular momentum just before the clay sticks?
What is the angular momentum of the bar+clay afterwards?