IIT JEE , AIEEE Forum - Physics , Mechanics

ramyani

Plz help me.

Q. Two particles each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of this system of particles is the same about any point taken as origin.

( Intro ex 9.3 of D C Pandey, volI, page 20 )

Plz give a detailed soln.

arpan1

here goes the solution

let us assume that one particle moves on the line y= mx + c

then as per ques, other particle moves along the line y = mx + d

then

angular momentum of first particle about origin= mvc / ( 1 +

m² )

angular momentum of second particle about origin = mvd / ( 1 +

m² )

clearly they are not equal

rate me

karthik2007

@arpan - the two lines that you have taken are not parallel. They are as shown by me in the figure.

madman

@arpan
parallel lines havent got opposite slopes
here is my solution
let the point be (x,y)
angular momentum of first particle = mvr1sin@ = mvp
angular momentum of second particle = mvr2sin@ = mvq
net angular momentum ( the angular momenta are in the same direction) = mvp+mvq =mvd (constant and independent of x and y)
you can also consider a point outside the lines
but in that case you will have to subtract them because they will be in opposite directions

madman

Re:D C Pandey Q frm rotational motion 2

anchitsaini

arpan1

@ madman and kartik

sorry, it was a typo . i hav corrected it (check it)

btw, madman why hav u specified by taking a point (x,y) and what is wrong with my solution

madman

@arpan
because i wanted to show that the angular momentum is the same irrespective of the point and x and y are variables
by origin they sont mean the point 0,0 but the point about which angular momentum is calculated

arpan1

see taking (0,0) or any other point hardly matters

u should get the correct ans

so what is wrong in my solution ????

anchitsaini

arey yaar
how does it "hardly matters"
you should get the correct answer but it would not be the correct proof
You have not read the qn properly
First of all it asks for angular momentum of the system and not of individual particles
Secondly it asks for angular momentum about any point which can be considered as origin but may not be 0,0

arpan1

thanks achit for ur reply

i do know that i am simplifying the proof by considering (0,0) than generalising it ...

but anyways it would yield the same ans

as for the angular momentum of the system , i hope it would b the addition of the angular momentum of the individual particles

can u state me where is wrong in my proof apart from the simplification

feynmann

I think the vector approach would be more simple to understand .

Suppose , wrt any origin chosen , the P.V. of the first and second particles be r1 and r2 .

Given that r1-r2= d + ap .............( 1) ( a is a scalar which depends on time ) ( p and -p their respective momentum )

Now denote the angular momentum of any particle by L

So , L1= r1 X p

L2= r2 X( -p) )

So L1+ L2 = ( r1-r2) X p = ( d+ a p ) Xp = d Xp = a constant ( independent of time )

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