How is Angular Momentum Calculated in the Bohr Model Using Polar Coordinates?

[dirtyhippy]

The angular momentum of the Bohr atom is ⃗L = ⃗r× ⃗p

Let the Oz axis of the unit vector ⃗uz be determined by the relation L = L⃗uz, in which L stands for the magnitude of the angular momentum vector ⃗L .
(Basically L is in Positive z axis)

We specify the position of the electron within its plane of movement
by the polar coordinates r = OM and θ = ∢(⃗ux,OM), where ⃗ux is the unit vector of the Ox axis. The unit vectors in polar coordinates are to be denoted ⃗ur and ⃗uθ. Calculate the absolute value L of the angular momentum as a function of m, r and dθ/dt.

What I've tried

Ive tried little things like dp=rdθ, to get ⃗v= ⃗w x ⃗r to ⃗v =dθ/dt x ⃗r
but really i have no idea how to bring that back ⃗L= ⃗r x ⃗p

Thanks for any help

 

[zachzach]

I'm not sure I understand your notation or work but here's my 2 cents.

\vec{r} \times \vec{p} = \abs{\vec{r}}\abs{\vec{p}}sin\theta and what is \theta?

 

[dirtyhippy]

I'm not sure I understand your notation or work
?

Sorry I should elaborate,

when there is a ⃗ in front of a letter it means vector, if not its a scalar

 

[dirtyhippy]

\vec{r} \times \vec{p} = \abs{\vec{r}}\abs{\vec{p}}sin\theta and what is \theta?

Theta is the angle between the x-axis and r (as the electron goes round).

Am I right in saying that it should be the magnitudes on the right hand side of your suggestion as opposed to vectors i.e.
\vec{r} \times \vec{p} = \abs{{r}}\abs{{p}}sin\theta
as then that may help me.

 

[zachzach]

yes about the magnitudes, couldn't get it in latex. Theta is the angle between the two vectors r and p.

 

[zachzach]

What is the question? To find the magnitude of L in terms of m, r and do/dt right? (o=theta)

 

[dirtyhippy]

aha got it, thankyou,

\vec{r} \times \vec{p} = \abs{{r}}\abs{{p}}sin\theta

using dx/dt = sin (x) dθ/dt and p =mv

we arrive at

L = r mv sin θ
L = r mv dθ/dt sin x
where x is actually ⃗ux

so we get L = r mv dθ/dt sin ( ⃗ux)


I think this is correct, thankyou zachzach

 

[zachzach]

aha got it, thankyou,

\vec{r} \times \vec{p} = \abs{{r}}\abs{{p}}sin\theta

using dx/dt = sin (x) dθ/dt and p =mv

we arrive at

L = r mv sin θ
L = r mv dθ/dt sin x
where x is actually ⃗ux

so we get L = r mv dθ/dt sin ( ⃗ux)I think this is correct, thankyou zachzach

I don't think so. Theta is a constant. It is the angle between r and p so the angle between r and v. Your answer is not in terms of what it asks for because you have a v in there. In fact your answer even has the wrong units. Also, is x an angle? Otherwise how why are you taking the sine of it?

 

[dirtyhippy]

ah balls your right, I've been looking at this question too long. I had to translate it from french as well which probably isn't helping. I'll take a nap and come back to it later.

but upto L=r m v sin theta is correct and v = r dθ/dt right, and theta being a constant means sin theta is also, so would it be

L=r^2 m dθ/dt sin theta, could i say theta is 90 ie sin theta is 1?
L=r^2 m dθ/dt ?

 

[zachzach]

That is what I got. Theta does equal 90 degrees always since it is traveling in a circle. V is tangential to the circle and r is the radial direction.

 

[dirtyhippy]

well in that case thankyou once again, much appreciated.

 

[zachzach]

no prob good luck.