How is the angular momentum related to x and y coordinates in SHM?

[thatguy14]

Homework Statement

Two-dimensional SHM: A particle undergoes simple harmonic motion in both the x and y directions
simultaneously. Its x and y coordinates are given by
x = asin(ωt)
y = bcos(ωt)

Show that the quantity x\dot{y}-y\dot{x} is also constant along the ellipse, where here the dot means the derivative with respect to time. Show that this quantity is related to the angular momentum of the system.

Homework Equations

L = mv x r

The Attempt at a Solution

Hi, so for the first part it is pretty simple and my answer is -abω, unless i made a dumb mistake which I don't think I did.

It's the second part that is giving me issues. How do I show that it is related to angular momentum? I tried doing this

L = \sqrt{L^{2}_{x}+L^{2}_{y}}

then L_{x} = m\frac{∂x}{∂t} x r

where r = \sqrt{x^{2}+ y^{2}}

and then plugging everything in. I was hoping all the cos and sin were going to cancel out but it got really huge and messy. I didn't think it was supposed to be that hard so can anyone tell me if I am going in the right direction or if there is something I am missing?

Thanks

 

[TSny]

Use the vector definition of angular momentum of a particle: $\vec{L} = \vec{r} \times \vec{p}= m\; \vec{r} \times \vec{v}$, where $\vec{r} = x\hat{i} + y\hat{j}$. (Note, the order of the cross product is important. Thus, $\vec{L} = m\;\vec{v} \times \vec{r}$ is not correct.)

 

[thatguy14]

Right whoops.

so then v = dx/dt i + dy/dt j

and when we cross them we get m(xdy/dt - ydx/dt) = L correct?

 

[TSny]

Yes. Good.