Angular momentum

  • #1
1,197
1
The problem is http://home.earthlink.net/~urban-xrisis/clip002.jpg [Broken].

I got a different answer from what my book tells me and I want to know why. Angular momentum is:
[tex]L=I \omega[/tex]
So... the moment of inertia for the block would be:
[tex]I=((M+m)l^2)[/tex]
since the rod has negligible mass, it will not have a moment of inertia.
[tex]\omega=v/r=v/l[/tex]
[tex]L=\frac{(M+m)l^2v}{l}[/tex]
[tex]L=vlM+vlm[/tex]

my text gives an answer of [tex]L=mvl[/tex]
I dont understand what I misunderstood
 
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Answers and Replies

  • #2
SpaceTiger
Staff Emeritus
Science Advisor
Gold Member
2,956
3
UrbanXrisis said:
my text gives an answer of [tex]L=mvl[/tex]
I dont understand what I misunderstood

Try again without assuming that the block/bullet combination are moving at the same speed as the original bullet. You'll need to consider conservation of linear momentum.
 

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