Angular Momentum of an object with respect to a moving point

[Tanya Sharma]

Homework Statement

Two particles A and B having equal masses m are rotating around a fixed point O with constant angular speed ω .A is connected to point O with a string of length L/2 whereas B is connected to point A with string of length L/2 .Find the angular momentum of B with respect to A about O.

O-------L/2-------A-------L/2--------B

Homework Equations

The Attempt at a Solution

There can be two approaches

1.We find angular momentum of B w.r.t O ,say L1 = mω²L.Then we find angular momentum of A w.r.t O ,say L2 = (mω²L)/4.Since angular momentum is a vector , angular momentum of B w.r.t A should be vector difference of L1 and L2 =(3/4)(mω²L)

2.We find relative velocity of B w.r.t A =ωL/2 ,i.e we have considered particle A to be at rest .Then we find shortest distance between point A and line of motion of B which is L/2 .
Now,angular momentum of A w.r.t B =(mω²L)/4

I feel approach 1 is the correct way of calculating angular momentum of a point with respect to a moving point ,but the correct answer is the one given by approach 2.

Which is the right way ?

 

[TSny]

Hello.

To see that method 1 doesn’t work, consider the case where the equal masses A and B are at the ends and the axis O is in the middle: A----------O----------B. The system revolves about O. Then A and B would have the same angular momentum vector about O. So, if you tried to get the angular momentum of B relative to A by subtracting the angular momenta of A and B about O, you would get zero. But that can’t be correct as B is moving relative to A.

For a general situation, you can see more formally why it doesn’t work by noting that $\vec{L}_{A/O} =m_A \vec{r}_{A/O} \times\vec{v}_{A/O}$. Similarly for particle B. If you subtract them, you can see that it's not possible in general to reduce it to $\vec{L}_{B/A} =m_B \vec{r}_{B/A} \times\vec{v}_{B/A}$

The reason you can subtract two velocity vectors to get a relative velocity is because of the relation between the position vectors: $\vec{r}_{B/A} = \vec{r}_{B/O} - \vec{r}_{A/O}$ which holds at each instant of time. Taking the time derivative gives the relative velocity formula.

 

[Tanya Sharma]

Hello TSny

Thank you very much for the explanation

So , I guess my misconception stemmed from the fact that I was treating Angular Momentum which is a vector in the same manner as we treat position vectors .ie $\vec{L}_{A/B} =\vec{L}_{A/O}+\vec{L}_{B/O}$ ,which is not the correct way. Instead I should have dealt Angular Momentum of A w.r.t as $\vec{L}_{B/A} =m_B \vec{r}_{B/A} \times\vec{v}_{B/A}$ .

Am I correct in assessing my mistake ?

 

[TSny]

Yes, I think that's right. There's no reason why angular momentum vectors would be related the same as position vectors.

 

[Tanya Sharma]

TSny...thanks once again