Angular Momentum Of A Particle

  • Thread starter mrshappy0
  • Start date
  • #1
99
0

Homework Statement



The position vector of a 1.0-kg particle is given by r = 4 i + 3 j + 3 k and its velocity by v = 5 i - 2 k. Find the magnitude of the particle's angular momentum in kg.m2/s.

Homework Equations



L=Iω

or

L=mvr

The Attempt at a Solution



This question is a real curve ball because I don't exactly know how to make sense of using this scalar system. I know that the object is moving in a linear path and that I am trying to find the angular momentum in relation to the origin. This means the angular momentum must be the same along the entire path. Thus the angular momentum would be mass*velocity*distance from the path of the particle to the parallel line extending from the origin. Since this problem is in the scalar form I don't really know what to do. I don't want an infraction so if you have any thoughts to help me move along that would be great. I have searched the internet to sources but nothing clarifies it well for me. I used hyperphysics.com which is usually great. Post any links that might help. Thanks.
 

Answers and Replies

  • #2
Doc Al
Mentor
45,029
1,326
Thus the angular momentum would be mass*velocity*distance from the path of the particle to the parallel line extending from the origin.
That's one way of looking at it. Consider the vector definition of angular momentum, since you're given the position and velocity vectors. See: http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html" [Broken]
 
Last edited by a moderator:
  • #3
99
0
So would it just be (4 i + 3 j + 3 k)*(5 i - 2 k)(mass). Completely lost. I am not sure how to deal with the i,j,k.
 
  • #4
Doc Al
Mentor
45,029
1,326
So would it just be (4 i + 3 j + 3 k)*(5 i - 2 k)(mass). Completely lost. I am not sure how to deal with the i,j,k.
You need the cross product (also called the vector product) of those vectors. The cross product can be defined in terms of unit vectors, as given here: http://hyperphysics.phy-astr.gsu.edu/hbase/vvec.html" [Broken]
 
Last edited by a moderator:
  • #5
99
0
I looked at both those pages and every page connected to them hours ago. I can't make sense of them. Super frusterated. Well the first page makes sense but finding the scalar product doesn't make sense.
 
  • #6
Doc Al
Mentor
45,029
1,326
I looked at both those pages and every page connected to them hours ago. I can't make sense of them. Super frusterated. Well the first page makes sense but finding the scalar product doesn't make sense.
You're not finding the scalar product here. (But the scalar product is the easy one, so I suspect you meant the vector product.)

Here's yet another presentation of the recipe for computing the cross product using unit vectors: http://en.wikipedia.org/wiki/Cross_product#Coordinate_notation"

Just follow the prescription and see what happens.
 
Last edited by a moderator:
  • #7
99
0
I got -5i+23j-k for r x v. Not sure where I go from here.
 
  • #8
Doc Al
Mentor
45,029
1,326
I got -5i+23j-k for r x v. Not sure where I go from here.
Double check that calculation.

Once you get the correct vector, find its magnitude just like any other vector.
 

Related Threads on Angular Momentum Of A Particle

  • Last Post
Replies
12
Views
10K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
358
Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
1
Views
4K
Replies
1
Views
3K
Replies
9
Views
852
Replies
1
Views
2K
Replies
7
Views
453
Top