# Angular momentum along a sloping line

1. Jul 18, 2015

### rpthomps

1. The problem statement, all variables and given/known data

A 1.0 kg particle is moving at a constant 3.5 m/s along the line y=0.62x +1.4, where x and y are in meters and where the motion is toward the positive x and y directions. Find its angular momentum about the origin

2. Attempt at a solution

$L=Iw\\\\L=myv\\\\L=(1)(0.62x+1.4)(3.5)$

Not sure what to do with x though. If I set x=0, this just evaluates the momentum at a point not over the line. The line is infinite, so I would have thought the momentum evaluates to infinity as well but the answer is 4.2

2. Jul 18, 2015

### Suraj M

Firstly, how did you get this formula? By the definition?
Rethink your substitution for y. What is $y$ by definition?

3. Jul 19, 2015

### rpthomps

You're right. There is a problem with my relationship.

The trig part doesn't seem to simplify to nicely though...

4. Jul 19, 2015

### haruspex

What is this point P you have chosen? Just consider the point where the trajectory crosses the y axis.

5. Jul 19, 2015

### rpthomps

Can I use that position because angular momentum will be conserved for the whole trip and thus will be the same along the path of the mass and the position you suggested is the simplest to calculate?

6. Jul 19, 2015

### haruspex

Yes.

7. Jul 19, 2015

### rpthomps

Then thank you sir for your help! Really appreciated.

8. Jul 21, 2015

### Suraj M

OP, since you've got the answer, it might help you in the future to know the formula for the perpendicular distance of a point from a line, which would simplify the calculation as there would be no angle involved in the calculation.
Do you happen to have a formula like that? if you did you'll get your d and hence answer would just be mvd.

9. Jul 26, 2015

Thank you.