Angular momentum along a sloping line

[rpthomps]

Homework Statement

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 origin2. 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

 

[Suraj M]

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

 

[rpthomps]

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

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

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

 

[haruspex]

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

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

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

 

[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?

 

[haruspex]

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?

Yes.

 

[rpthomps]

Then thank you sir for your help! Really appreciated.

 

[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.

 

[rpthomps]

Thank you.