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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

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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