Angular momentum of a pulley system

[undividable]

Homework Statement

it is given in the image i uploaded

Homework Equations

L^→=r^→ × p^→[/B]

The Attempt at a Solution

If the angular momentum of a particle is the cross product of the position vector of the particle from the axis and its linear momentum, how can the angular momentum of m1 and m2 be m1vR and m2vR respectively, R is not the position vector, it's just the radius of the pulley. The position vector should be a vector that starts at the axis and ends at the center of mass of m1 or m2, R is not even equal in magnitude to the position vector of m2 or m1.

 

[kuruman]

Assume that the entire mass of the sphere and the block is concentrated at one and the other end of the string. Does that help?

 

[DrClaude]

Have you tried calculating the angular momentum?

 

[undividable]

Have you tried calculating the angular momentum?

i have, but the problem says that in the analyzing part it says that m1vR and m2vR are the angular momentum for m1 and m2 respectively, and i can not understand why we use R, that is not the position vector.

 

[undividable]

Assume that the entire mass of the sphere and the block is concentrated at one and the other end of the string. Does that help?

i did, and my question remains, why does the text in the image say that m1vR and m2vR are the angular momentum for m1 and m2 respectively, R is not the position vector, even if we consider the center of mass of m1 and m2 as what the position vector is referring to

 

[DrClaude]

i have, but the problem says that in the analyzing part it says that m1vR and m2vR are the angular momentum for m1 and m2 respectively, and i can not understand why we use R, that is not the position vector.

That tells me you haven't actually calculated $\vec{L}$. You have masses moving is straight lines, so it shouldn't be complicated to figure out $\vec{r}$ and $\vec{p}$, and hence $\vec{L}$.

 

[undividable]

That tells me you haven't actually calculated $\vec{L}$. You have masses moving is straight lines, so it shouldn't be complicated to figure out $\vec{r}$ and $\vec{p}$, and hence $\vec{L}$.

well, i can't figure out r^→

 

[DrClaude]

well, i can't figure out r^→

If you choose your coordinate system cleverly, you will find that there is one vector components $\vec{r}$ that is time-dependent, so you don't actually know what it is, but the other two are fixed, and the one you don't know won't matter.

 

[undividable]

If you choose your coordinate system cleverly, you will find that there is one vector components $\vec{r}$ that is time-dependent, so you don't actually know what it is, but the other two are fixed, and the one you don't know won't matter.

so the component that is constant is equal to R and is perpendicular to the momentum vector, and the other vector component is time-dependent and parallel to the momentum, so the vector product will equal a vector with components R and 0?

 

[kuruman]

i have, but the problem says that in the analyzing part it says that m1vR and m2vR are the angular momentum for m1 and m2 respectively, and i can not understand why we use R, that is not the position vector.

Study the drawing. The angular momentum of the mass on the surface is $\vec{L}_2 = \vec{r}_2 \times \vec{p}_2$. What is the magnitude of the cross product? Does it depend on the magnitude of $\vec{r}_2$? This is another way of saying what @DrClaude said.