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Angular momentum of a particle in Classical Mechanics 
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#1
May2711, 05:15 PM

PF Gold
P: 3,188

1. The problem statement, all variables and given/known data
Calculate the Cartesian expressions and the value of the modulus of the angular momentum in cylindrical coordinates of a particle whose coordinates are [itex](r, \phi, z)[/itex]. 2. Relevant equations [itex]L=TV[/itex], [itex]\vec P = \sum _i ^3 \frac{\partial L}{\partial \vec {\dot q_i}}[/itex], [itex]\vec M = \sum _i^3 \vec r_i \times \vec P_i [/itex]. 3. The attempt at a solution Not sure what they mean with Cartesian expressions. The position of such a particle in Cartesian coordinates is [itex](\sqrt {x^2+y^2},\arcsin \left ( \frac{y}{\sqrt{x^2+y^2}} \right ),z)[/itex] if it's what they ask for. For the Lagrangian, [itex]V=0[/itex] so [itex]L=\frac{m}{2}v^2[/itex]. I've calculated [itex]v^2[/itex] to be worth [itex]\dot r^2 + r^2 \dot \phi ^2 + \dot z ^2[/itex]. This gave me [itex]\vec P =(m \dot r, m r^2 \dot \phi , m \dot z)[/itex]. As for [itex]\vec M[/itex] I'm not so sure. I took [itex]\vec r[/itex] as [itex](r, \phi, z)[/itex] but this doesn't really make sense to me. Anyway this gave me [itex]\vec M=m(\dot z \phi z r^2 \dot \phi) \hat i +m(\dot r zr\dot z) \hat j +m(r^3 \dot \phi  \phi \dot r) \hat k[/itex]. Now I have to take the square of each component, sum them all and take the square root of it. But I'm not confident in what I've done so far. Could you please enlighten me? 


#2
May2711, 06:12 PM

P: 205

r = (x,y,0)
I would convert L into Cartesian and work with that 


#3
May2711, 06:39 PM

PF Gold
P: 3,188

First of all thanks for your help,
About converting the Lagrangian in Cartesian, since it's simply [itex]\frac{m v^2}{2}[/itex], [itex]v^2[/itex] would be worth [itex]\dot x^2 + \dot y^2 + \dot z ^2[/itex]. 


#4
May2811, 07:29 AM

P: 205

Angular momentum of a particle in Classical Mechanics
I don't assume it is in the xy plane from what you have described the thing is rotating around the z axis which would mean it's r vector is only given by x and y. Furthermore you already gave me that expression for r when you said [tex] r = \sqrt{x^2 + y^2} [/tex]. As for the Cartesian expression, isn't that what the question is asking you. If you want you could take
[tex] \vec{r} = (rcos \theta , rsin \theta , 0 ) [/tex] then you can call that the cartesian expression 


#5
May2811, 05:19 PM

PF Gold
P: 3,188

My r isn't the modulus of [itex]\vec r[/itex] but the modulus of the projection of [itex]\vec r[/itex] into the xy plane. In wikipedia (http://en.wikipedia.org/wiki/File:Coord_system_CY_1.svg) this would be the [itex]\rho[/itex] coordinate. I'm not told the particle is even moving. I guess I want to reach an expression for the angular momentum that depends on the derivative of some coordinate(s) with respect to time and if by chance it/they appear(s) to be non zero then the particle is moving/rotating but since I'm not told anything about a motion I cannot assume these are non zero and hence the particle moves. 


#6
May2811, 10:09 PM

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PF Gold
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I think the problem wants you to calculate, for example,
[tex]L_x = yp_zzp_y = (r\sin\phi)(m\dot{z})z(m\dot{r}\sin\phi+mr\dot{\phi}\cos\phi)[/tex] Then once you have all the components of the angular momentum L, find its modulus. 


#7
May2811, 10:54 PM

PF Gold
P: 3,188

However it means I made an error with [itex]\vec P[/itex]. Your [itex]P_y=m\dot{r}\sin\phi+mr\dot{\phi}\cos\phi[/itex] while mine is worth [itex]mr^2 \phi[/itex]. My Lagrangian is worth [itex]\frac{m v^2}{2}[/itex] with [itex]v^2=\dot r^2 + r^2 \dot \phi ^2 + \dot z ^2[/itex]. I calculated [itex]\vec P[/itex] as [itex](\frac{\partial L}{\partial \dot r},\frac{\partial L}{\partial \dot \phi},\frac{\partial L}{\partial \dot z})[/itex]. I need a correct [itex]\vec P[/itex] in order to get a correct [itex]\vec M[/itex] (or your [itex]\vec L[/itex] :) ) I really do not see where I went wrong with my [itex]\vec P[/itex]! How did you get your [itex]\vec P[/itex]? 


#8
May2811, 11:17 PM

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The velocity is [itex]\vec{v} = v_r \hat{r} + v_\phi \hat{\phi} + v_z \hat{z}[/itex] where [itex]v_r = \dot{r}[/itex], [itex]v_\phi = r\dot{\phi}[/itex], and [itex]v_z = \dot{z}[/itex]. You should recognize [itex]v_\phi[/itex] as being the tangential velocity of an object, if it were rotating about an axis at a distance r with an angular velocity of [itex]\dot{\phi}[/itex]. If you were to draw a picture, you'd see both [itex]v_r[/itex] and [itex]v_\phi[/itex] generally have components in the x and y directions, so you need to add both contributions from each to get components of velocity in the x and y direction.
The components of your [itex]\vec{P}=(\frac{\partial L}{\partial \dot r},\frac{\partial L}{\partial \dot \phi},\frac{\partial L}{\partial \dot z})[/itex] are the momenta conjugate to [itex](r,\phi,z)[/itex]. They're not the [itex]P_i[/itex]'s that appear in your definition for [itex]\vec{M}[/itex], which are the Cartesian components of momentum. There's a decent picture on http://www.realworldphysicsproble...armotion.html about halfway down, under "Curvilinear Motion In Polar Coordinates." 


#9
May2911, 11:12 AM

PF Gold
P: 3,188

Thanks once again for the help!
Oh I didn't know about my error for the use of my P into the definition of M... So I have to translate my P into a Cartesian one? I don't think I have to change my [itex]\dot \phi[/itex] by [itex]\frac{d}{dt} \arcsin \left ( \frac{y}{\sqrt{x^2+y^2}} \right )[/itex] which seems a real mess. How can I change my P into yours? 


#10
May2911, 01:14 PM

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#11
May2911, 02:20 PM

PF Gold
P: 3,188

Well I'm a bit lazy to simplify the final expression of the angular momentum but it's worth [itex]\vec L=m \sqrt{[r\sin( \phi )\dot zz(\dot r \sin (\phi )+r\dot \phi \cos (\phi))]^2+[z(\dot r \cos (\phi)  \sin (\phi ))r \cos (\phi )\dot z]^2+[r\cos (\phi) (\dot r \sin (\phi)+r \dot \phi \cos (\phi ))r\sin (\phi )(\dot r \cos (\phi )\sin (\phi) )]^2}[/itex]. So basically [itex]\vec L = (x,y,z) \times (P_x, P_y, P_z)[/itex] okay. I have only 1 remaining question... why if I take [itex]\vec P =m \vec v[/itex] I get [itex]\vec P =m(\dot r \hat r + r \dot \phi \hat \phi+\dot z \hat z)[/itex]. While if I calculate [itex]\vec P = \sum _i ^3 \frac{\partial L}{\partial \vec {\dot q_i}}[/itex] I get [itex]\vec P =\vec P =(m \dot r, m r^2 \dot \phi , m \dot z)[/itex] which slightly differs from [itex]\vec P =m\vec v[/itex]. In fact it only differs in [itex]P_\phi[/itex]. When I derive the Lagrangian with respect to [itex]\dot \phi[/itex] I get [itex]m r^2 \dot \phi[/itex]. While the y component of [itex]m \vec v[/itex] is [itex]mr \dot \phi[/itex]. This is really bothering me... I'd appreciate if you could explain me what's wrong there. Thanks for all so far :) 


#12
May2911, 03:39 PM

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I just realized I gave you a roundabout way of getting the Cartesian components of velocity. You could have instead differentiated [itex]x = r\cos\phi[/itex] to get [itex]v_x = \dot{r}\cos\phir\sin\phi\dot{\phi}[/itex] and so on.
The conjugate momenta don't necessarily correspond to components of a vector. Note that [itex]\partial L/\partial \dot{r}[/itex] and [itex]\partial L/\partial \dot{\phi}[/itex] have different units. The first momentum is the linear momentum in the radial direction, while the second momentum is an angular momentum, specifically L_{z}, since varying [itex]\phi[/itex] results in rotation about the z axis. If you simplify your expression for L_{z}, you should get [itex]r^2\dot{\phi}[/itex], which is equal to [itex]\partial L/\partial \dot{\phi}[/itex]. 


#13
May2911, 05:48 PM

PF Gold
P: 3,188

And yes, differentiating x with respect to time is kind of easier than finding the unit vectors of cylindrical coordinates in function of the one of Cartesian coordinates. I'll try to keep all this in mind. Problem solved. 


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