# Angular Speed, Glob and Rod

## 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?

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## Answers and Replies

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$$L_i=RXP=mvb=1.6929$$
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.

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

Thanks, I got it.

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