Register to reply

Angular momentum cross product - Please help!

by thehiggsboson
Tags: angular momentum, cross product
Share this thread:
thehiggsboson
#1
Dec1-12, 02:28 PM
P: 4
1. The problem statement, all variables and given/known data

A 1.47kg particle moves in the xy plane with a velocity of v = (4.59i - 3.28j)m/s. Determine the magnitude of the particle's angular momentum when its position vector is r = (1.35i + 2.57j)m.


2. Relevant equations

p = mv

L = r x p (the x is supposed to be a cross product and not a variable)

L = r x mv

3. The attempt at a solution

First I scaled the velocity vector: v = (4.59i - 3.28j)m/s by the mass, 1.47 kg, to get a new momentum vector (6.75i - 4.82j)kg*m/s.

Then I took the cross product of the r vector with the new momentum vector:
(1.35i + 2.57j)m x (6.75i - 4.82j)kg*m/s (I let a=1.35, b=2.57, c=0, d=6.75, e=-4.82, and f=0, the got the k vector cross product by doing k=ae-bd)

The answer I got was -23.9 kg*m^2/s, which wasn't right.

What did I do wrong? Am I even anywhere near the correct solution/answer?
Phys.Org News Partner Science news on Phys.org
Mysterious source of ozone-depleting chemical baffles NASA
Water leads to chemical that gunks up biofuels production
How lizards regenerate their tails: Researchers discover genetic 'recipe'
gneill
#2
Dec1-12, 03:16 PM
Mentor
P: 11,676
Quote Quote by thehiggsboson View Post
1. The problem statement, all variables and given/known data

A 1.47kg particle moves in the xy plane with a velocity of v = (4.59i - 3.28j)m/s. Determine the magnitude of the particle's angular momentum when its position vector is r = (1.35i + 2.57j)m.


2. Relevant equations

p = mv

L = r x p (the x is supposed to be a cross product and not a variable)

L = r x mv

3. The attempt at a solution

First I scaled the velocity vector: v = (4.59i - 3.28j)m/s by the mass, 1.47 kg, to get a new momentum vector (6.75i - 4.82j)kg*m/s.

Then I took the cross product of the r vector with the new momentum vector:
(1.35i + 2.57j)m x (6.75i - 4.82j)kg*m/s (I let a=1.35, b=2.57, c=0, d=6.75, e=-4.82, and f=0, the got the k vector cross product by doing k=ae-bd)

The answer I got was -23.9 kg*m^2/s, which wasn't right.

What did I do wrong? Am I even anywhere near the correct solution/answer?
Did you specify the direction of the angular momentum? (It's a pseudo-vector, so it has a magnitude and direction).
thehiggsboson
#3
Dec1-12, 03:33 PM
P: 4
I didn't, but I just went back and tried it and it picked up the k as being part of the unit. So, it's not that, I don't think. The question also just asks for magnitude.

gneill
#4
Dec1-12, 03:36 PM
Mentor
P: 11,676
Angular momentum cross product - Please help!

Quote Quote by thehiggsboson View Post
I didn't, but I just went back and tried it and it picked up the k as being part of the unit. So, it's not that, I don't think. The question also just asks for magnitude.
Okay, just the magnitude. What answer did you type in for that?
thehiggsboson
#5
Dec1-12, 03:38 PM
P: 4
-23.9 kg*m^2/s
gneill
#6
Dec1-12, 03:39 PM
Mentor
P: 11,676
Are magnitudes ever negative?
thehiggsboson
#7
Dec1-12, 03:41 PM
P: 4
Nope. Wow, I feel a little stupid now. Thanks!
gneill
#8
Dec1-12, 03:44 PM
Mentor
P: 11,676
Quote Quote by thehiggsboson View Post
Nope. Wow, I feel a little stupid now. Thanks!
Heh. We've all been there, done that, got the T-shirt. Cheers.


Register to reply

Related Discussions
Given one cross product, find another cross product Calculus & Beyond Homework 1
Deriving cross product and dot product, stuck at beginning. Calculus & Beyond Homework 8
Angle between 2 vectors using 1) Dot product and 2) cross product gives diff. answer? Calculus & Beyond Homework 8
Cross product and dot product of forces expressed as complex numbers Introductory Physics Homework 4
Angular Momentum, Clebsch Gordan and Inner Product General Physics 6