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Angular Momentum of each particle about the origin

  • Thread starter joemama69
  • Start date
  • #1
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Homework Statement


Determine the Angular Momentum of each particle about the origin

Point A Located @ (-8,12), 6kg,4m/s, headed 60degrees below the horizon to the right
Point B (2,1.5),4kg, 6m/s, headed 30 degrees above the horizon to the right
Point C (6,-2), 2kg, 2.6m/s, headed on a 5,12,13 angled to the left and down


Homework Equations



L=r x p
p=mv


The Attempt at a Solution



A. -8i+12j X 24(-i-8.66j)=
B. 2i+1.5j X 24(i+.5j)=
c. 6i-2j X 5.2(-12i-5j)=

If this is right i dont know how to calculate it
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
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Determine the Angular Momentum of each particle about the origin

Point A Located @ (-8,12), 6kg,4m/s, headed 60degrees below the horizon to the right
Point B (2,1.5),4kg, 6m/s, headed 30 degrees above the horizon to the right
Point C (6,-2), 2kg, 2.6m/s, headed on a 5,12,13 angled to the left and down

L=r x p
p=mv

A. -8i+12j X 24(-i-8.66j)=
B. 2i+1.5j X 24(i+.5j)=
c. 6i-2j X 5.2(-12i-5j)=

If this is right i dont know how to calculate it
Not quite right. To the right is +i isn't it?
A. -8i+12j X 24(-i-8.66j)=
should be
A. (-8i + 12j) X 24*(.5i - .866j)=

B. 2i+1.5j X 24(i+.5j)=
should be
B. (2i + 1.5j) X 24*(.866i + .5j)=

c. 6i-2j X 5.2(-12i-5j)=
should be
c. (6i - 2j) X 5.2*(-12i - 5j -13k) =

To take the cross product ...
http://en.wikipedia.org/wiki/Cross_product#Computing_the_cross_product
 
Last edited:
  • #3
rl.bhat
Homework Helper
4,433
5
(-8i + 12j) X 24*(.5i - 8.66j)

It should be (-8i + 12j) X 24*(.5i - .866j)
 
  • #4
LowlyPion
Homework Helper
3,090
4
(-8i + 12j) X 24*(.5i - 8.66j)

It should be (-8i + 12j) X 24*(.5i - .866j)
Indeed it should. Thanks for the catch.
 
  • #5
399
0
ok i see my mistakes.

just to clarify @ point C

13 is not the k value, it is the third side of the triangle/hipotinuse
 
  • #6
399
0
Ok so here what i got

A. (-8i+12j)X(12i-20.784j)=22.72k
B. (2i+1.5j)X(2.784i+12j)=-7.176k
C. (6i-2j)X(62.4i-26j)=280k
 
  • #7
LowlyPion
Homework Helper
3,090
4
Ok so here what i got

A. (-8i+12j)X(12i-20.784j)=22.72k
B. (2i+1.5j)X(2.784i+12j)=-7.176k
C. (6i-2j)X(62.4i-26j)=280k
A. (-8 i + 12 j) X (12 i - 20.784 j)= 22.272 k

C. (6i-2j)X(62.4i-26j)=280k
Looking at the original direction again I think it should be -26 i -62.4 j. This yields a different result.
 
  • #8
399
0
Nope the triangle is not oriented like that


the bottom is 12 or -12i, the side is on the right and up -5j.
 
  • #9
LowlyPion
Homework Helper
3,090
4
Nope the triangle is not oriented like that


the bottom is 12 or -12i, the side is on the right and up -5j.
OK. I can't see the picture, so going with that ... so using -62.4 i -26 j
shouldn't it be -156 - 124.8 = -280.8 k
 

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