Angular Momentum of 1.80-kg Particle in xy Plane

[AEfly]

Homework Statement

A 1.80-kg particle moves in the xy plane with a velocity of = (4.30i − 3.70j ) m/s. Determine the angular momentum of the particle about the origin when its position vector is = (1.50i + 2.20j ) m.

The Attempt at a Solution

I get an answer of -29.016 Kg m2 per second using a known method but it comes out as incorrect. An ideas?

 

[eczeno]

what did you try? I get about -27 when I work it out.

 

[AlexChandler]

The angular momentum is

\vec{L} = \vec{r} \times \vec{p}

where p is the momentum

we have

\vec{r} = (1.50 \hat{i} + 2.20 \hat{j} + 0 \hat{k} ) \textnormal{m}

and

\vec{p} = m ( \vec{v} ) = m ( 4.30 \hat{i} - 3.70 \hat{j} + 0 \hat{k} ) \textnormal{ m / s }

then

\vec{L} = \vec{r} \times \vec{p} = (1.50 \hat{i} + 2.20 \hat{j} + 0 \hat{k} ) \textnormal{m} \times m ( 4.30 \hat{i} - 3.70 \hat{j} + 0 \hat{k} ) \textnormal{ m / s }

This is

\vec{L} = - 27.018 \hat{k} \textnormal{ kg m^2 / s }

 

[AlexChandler]

This is

\vec{L} = - 27.018 \hat{k} \textnormal{ kg m^2 / s }

Ignore everything after kg m^2 / s ... I don't know what this has happened... I tried to fix it. Why is this happening? I am new to latex by the way :D

 

[eczeno]

that's wild. i have never seen that, it is like the latex abyss.