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Angular Speed, Glob and Rod

  • #1

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



[tex]L_i=RXP=I\omega[/tex]

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

Parallel Axis Theorem

The Attempt at a Solution



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

[tex]L_i=RXP=mvb=1.6929[/tex]

[tex]L_f=I_{glob}\omega+I_{rod}\omega[/tex]

[tex]\delta=distance from old cm to new cm[/tex]

[tex]\delta=\frac{mb}{m+M}=.1633 m[/tex]

[tex]I_{glob}=m(b-\delta)^2[/tex]

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

[tex]L_i=L_f[/tex]

I get 8.15 rad/s which is wrong, the right answer is 5.743. What did I do wrong?
 
Last edited by a moderator:

Answers and Replies

  • #2
Doc Al
Mentor
44,912
1,170
[tex]L_i=RXP=mvb=1.6929[/tex]
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.
 
  • #3
So I should have [tex]L_i=mv(b-\delta)[/tex] ?

Thanks, I got it.
 
Last edited:

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