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Angular Momentum of an object with respect to a moving point

  1. Dec 20, 2012 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations



    3. 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 ?
     
  2. jcsd
  3. Dec 20, 2012 #2

    TSny

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    Gold Member

    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.
     
  4. Dec 20, 2012 #3
    Hello TSny

    Thank you very much for the explaination :smile:

    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 [itex] \vec{L}_{A/B} =\vec{L}_{A/O}+\vec{L}_{B/O} [/itex] ,which is not the correct way. Instead I should have dealt Angular Momentum of A w.r.t as [itex]\vec{L}_{B/A} =m_B \vec{r}_{B/A} \times\vec{v}_{B/A}[/itex] .

    Am I correct in assessing my mistake ?
     
  5. Dec 20, 2012 #4

    TSny

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    Yes, I think that's right. There's no reason why angular momentum vectors would be related the same as position vectors.
     
  6. Dec 20, 2012 #5
    TSny...thanks once again
     
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