Atwoods machine and angular momentum

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demonelite123
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so given atwood's machine, a pulley with mass M and radius R, a massless string connected to two blocks with masses m1 and m2 with m1 > m2 and the string does not slip. calculate the angular momentum of the system and use it to find the angular acceleration of the pulley and the linear acceleration of the blocks.

i noticed the book calculate the angular momentum of the system as Iw + m1vR + m2vR. i undestand the pulley having angular momentum Iw but i don't understand why the angular momentum of the blocks being m1vR and m2vR. for a system of particles i know that its just the momentum mv of the particle times the distance of the particle from the point we choose to find the angular momentum with respect to.

im confused because how can the angular momentum of the blocks be m1vR and m2vR when the blocks are connected to strings and they are farther away from the center of the pulley than R? i think I'm confused on which point to take the angular momentum respect to. when i pick the point to be the center of the pulley, it doesn't seem to work out?
 
on Phys.org
The point is the center of the pulley. That is for sure.

Angular momentum = r cross mv which means rmvsin(theta). Here r is the distance between the center and m1 and theta is angle between r and v. When you will find (by diagram) rsin(theta) you will get R. So the angular momentum = m1vR.
- ashishsinghal
 
You're finding the angular momentum about the axis of the pulley. What's important isn't the displacement vector (distance) that m1 is from the axis of the pulley, but the component of the displacement vector that's perpendicular to the momentum vector, m1v1.
 
yeah, you can say so
 
you can also say that velocity is not important, velocity perpendicular to displacement vector is important. Mainly, mvrsin(theta) is important, you can approach it in any way.
-ashishsinghal
 
ah i see. i forgot about the perpendicular component of the distance. thanks guys!