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Angular momentum - i + j

  • Thread starter mybrohshi5
  • Start date
  • #1
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Homework Statement



A particle has a mass of 2kg and moves in the xy plane with a constant speed of 3m/s along the direction of the vector r = i + j. What is the particle's angular momentum relative to the point (0,5) meters?

Homework Equations



L = rmv

L = mvrsin(theta)

The Attempt at a Solution



I am studying for my exam and i am a little confused about this problem.

I know the answer is 21 kgm2/s

and the work to get that answer is (2kg)(3m/s)(5m)sin(45) = 21

So i get where the mass (2kg) and the velocity (3m/s) go in the equation but i dont get why the 5m is in there and how would i know to use the angle 45 degrees?

Thanks for any help on clarifying this up for me :)
 

Answers and Replies

  • #2
88
1
From the vector r=i+j , you can see the particle at some time will be at point (0,0). Here, it is 5 meters from the point (0,5). The angle is that between the line that connects the particle and the point and the movement vector of the particle. The line goes straight up, and the movement vector is at 45 degrees away from this.
You should really always make a drawing for these kinds of questions, that makes it really easy :)
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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Hi mybrohshi5! :smile:
A particle has a mass of 2kg and moves in the xy plane with a constant speed of 3m/s along the direction of the vector r = i + j. What is the particle's angular momentum relative to the point (0,5) meters?

L = rmv

L = mvrsin(theta)

So i get where the mass (2kg) and the velocity (3m/s) go in the equation but i dont get why the 5m is in there and how would i know to use the angle 45 degrees?

Thanks for any help on clarifying this up for me :)
(btw, there's a third formula for angular momentum … L = r x mv, using the cross product :wink:)

45º is the angle of the vector i + j.

5sin45º is the "lever arm", or the distance from the point (0,5) to the line of the momentum (draw it, and you'll see). :smile:
 
Last edited:
  • #4
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try the cross product of the position and the velocity vectors
 
  • #5
365
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Ok i think i get it now. i understand the lever arm part so if i draw it again i think i will get it :)

Thanks everyone for the fast replies
 
  • #6
365
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I think i am not drawing it right because from my picture my lever arm isn't 5sin(45)

So i have a normal x y plane.

i have a line going through the first quadrant at 45 degrees (i think this is r)

then i have a dot at the point (0,5) and then a line from (0,5) to the line r and i labeled this L.

(so to try and make it clear for you guys i have a line from the origin to (0,5) then a line directly to the right labeled L and then a diagonal line from the end of L back to the origin make an upside down 45 degree triangle)

So what am i doing wrong in my picture?

Thanks :)
 
  • #7
tiny-tim
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Hi mybrohshi5! :wink:
(so to try and make it clear for you guys i have a line from the origin to (0,5) then a line directly to the right labeled L and then a diagonal line from the end of L back to the origin make an upside down 45 degree triangle)
I think you're drawing L (btw, don't use L … L is for angular momentum!! :rolleyes:) "horizontally", but it should be the shortest distance from (0,5) to the line of momentum. :smile:
 
  • #8
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I guess i am confused on what the line of momentum is then in the picture.
 
  • #9
tiny-tim
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The line of momentum is the up-right diagonal line through the origin.

The "lever arm" is the down-right diagonal line from (0,5). :smile:
 
  • #10
365
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Ok i got that drawn out now, but i just have one last question.

If the "lever arm" is a down-right diagonal line from (0,5) to the L how does this make a 45, 45, 90 triangle?

I thought the "lever arm" would have to be horizontal from (0,5) to L in order to make a 45 degree angle triangle...
 
  • #11
365
0
AHH never mind i figured it out :)

Thanks for the help tim :)
 

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