Angular momentum of particle about the origin

[Krushnaraj Pandya]

Homework Statement

A particle (5 kg) moves with constant velocity 2 m/s along the straight line 2y=3x+4, the angular momentum of the particle about origin is?

Homework Equations

L=r x p

The Attempt at a Solution

For a 2d problem we take the component of velocity perpendicular to the point about which we want to find the momentum and multiply it with the perpendicular distance. Here slope of the line gives tanθ=3/2, p=mv=10 and L=mvrcosθ, mvcosθ equals 20/√13 but isn't the distance r (here r=the y coordinate of the particle) continuously increasing? I get the correct answer for r=2 though, I don't understand how. I'd really appreciate some help, thank you

 

[Orodruin]

$\vec r \times \vec p$ only depends on the part of $\vec r$ that is orthogonal to $\vec p$.

 

[Krushnaraj Pandya]

$\vec r \times \vec p$ only depends on the part of $\vec r$ that is orthogonal to $\vec p$.

I know only the very basics of vectors, (its yet to be reached in our mathematics coursework). I suppose the part of r that's perpendicular to momentum is what you mean but that part seems to be increasing with time as well

 

[Orodruin]

but that part seems to be increasing with time as well

Why do you think so? (It is not correct)

 

[Orodruin]

Here is a quick drawing of the situation.
The red vector is always the part of the position vector orthogonal to p. The blue vectors are different position vectors for the part let moving along the black line and the magenta vector its momentum. The green vectors are the components of the position vectors not perpendicular to p.

 

[Krushnaraj Pandya]

View attachment 232643
Here is a quick drawing of the situation.
The red vector is always the part of the position vector orthogonal to p. The blue vectors are different position vectors for the part let moving along the black line and the magenta vector its momentum. The green vectors are the components of the position vectors not perpendicular to p.

Ohh, I understand really well where I was thinking wrong. The diagram helped a lot. Thank you :D
The fact that people on this website take out so much time to help students like me amazes me, so I've thought about donating to the site from my savings after my exams this year are over. can you direct me to where I can do that?

 

[Orodruin]

The fact that people on this website take out so much time to help students like me amazes me, so I've thought about donating to the site from my savings after my exams this year are over. can you direct me to where I can do that?

If you want to support PF I would suggest a gold membership. It supports PF and you get something out of it. See also
https://www.physicsforums.com/threads/easy-ways-you-can-support-physics-forums.813856/

 

[Krushnaraj Pandya]

If you want to support PF I would suggest a gold membership. It supports PF and you get something out of it. See also
https://www.physicsforums.com/threads/easy-ways-you-can-support-physics-forums.813856/

Alright, thank you very much