UrbanXrisis
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The problem is http://home.earthlink.net/~urban-xrisis/clip002.jpg .
I got a different answer from what my book tells me and I want to know why. Angular momentum is:
L=I \omega
So... the moment of inertia for the block would be:
I=((M+m)l^2)
since the rod has negligible mass, it will not have a moment of inertia.
\omega=v/r=v/l
L=\frac{(M+m)l^2v}{l}
L=vlM+vlm
my text gives an answer of L=mvl
I don't understand what I misunderstood
I got a different answer from what my book tells me and I want to know why. Angular momentum is:
L=I \omega
So... the moment of inertia for the block would be:
I=((M+m)l^2)
since the rod has negligible mass, it will not have a moment of inertia.
\omega=v/r=v/l
L=\frac{(M+m)l^2v}{l}
L=vlM+vlm
my text gives an answer of L=mvl
I don't understand what I misunderstood
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