Calculating Angular Speed of a Bar/Glob System After Impact

[americanforest]

Homework Statement

http://loncapa2.physics.sc.edu/res/msu/physicslib/msuphysicslib/21_Rot3_AngMom_Roll/graphics/prob21a_bar.gif

On a frictionless table, a glob of clay of mass 0.38 kg strikes a bar of mass 0.90 kg perpendicularly at a point 0.55 m from the center of the bar and sticks to it. If the bar is 1.30 m long and the clay is moving at 8.1 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

L_i=RXP=I\omega

cm_[new]=\frac{all m*d}{total mass}

Parallel Axis Theorem

The Attempt at a Solution

M is mass of rod and m is mass of glob.

L_i=RXP=mvb=1.6929

L_f=I_{glob}\omega+I_{rod}\omega

\delta=distance from old cm to new cm

\delta=\frac{mb}{m+M}=.1633 m

I_{glob}=m(b-\delta)^2

I_{rod}=\frac{M(L)^2}{12}+M\delta^2

L_i=L_f

I get 8.15 rad/s which is wrong, the right answer is 5.743. What did I do wrong?

 

[Doc Al]

L_i=RXP=mvb=1.6929

This is the angular momentum of the blob about the center of the bar. What you need is the angular momentum about the center of mass of the system--that is the quantity that is conserved.

 

[americanforest]

So I should have L_i=mv(b-\delta) ?

Thanks, I got it.