Angular momentum of time dependent particle motion

AI Thread Summary
The discussion focuses on calculating the angular momentum of a particle moving in a circle with radius R and constant speed v, specifically about a point P located at (-R, 0). The initial attempt involved using the formula for angular momentum, but the user realized their calculations were incorrect as they did not account for the vector nature of angular momentum. Key insights included the need to distinguish between angular momentum about the origin and point P, and the importance of correctly interpreting the cross product of position and momentum vectors. Ultimately, the correct expression for angular momentum was derived as L = (mvR) cos((vt/R) + 1), highlighting the significance of proper vector analysis in the solution. This illustrates the complexities involved in angular momentum calculations for time-dependent particle motion.
Schulze
Messages
18
Reaction score
0

Homework Statement


A particle of mass m moves in a circle of
radius R at a constant speed v. Assume: The
motion begins from the point Q, which has
coordinates (R, 0).
Determine the angular momentum of the
particle about point P, which has coordinates
(−R, 0) as a function of time.

The answer choices can be found at: http://imgur.com/GxVLhGb


Homework Equations


v = Rω
L = R x p = Rpsin(θr,p) = Rmv(sinθr,p)
θf = θi + (ωit)
θf = θi + (vt/R)

The Attempt at a Solution


L1 at position Q = R x p = Rpsin(θr,p) = Rmv(sinθr,p)

L2 at position P = R x p
L2 at position P = Rpsin(-θr,p)
L2 at position P = Rmv(-sinθr,p)
L2 at position P = Rmv(sin((vt/R)+∏)))

However all answers have Rmv(____ + 1) factor. Which I do not have. Therefore I have reason to believe that my answer is incorrect.

My guess is that the answer is Rmv(sin((vt/R)+∏))+1)
 
Physics news on Phys.org
1. Make a drawing
2. Realize that L = r X p is NOT Rmv(sinθr,p) but that ##\vec L= \vec r(t) \times \vec p(t)##: all the answers are vectors!
3. Make a distinction between angular momentum about the origin and angular momentum about point (-R, 0)
 
I did all those things but I thought that this site didn't want me to post a poorly drawn sketch. Furthermore, I am not very good at using Latex to make my equations look nice and bold and whatnot.

Nor am I very efficient at using this yet so bare with me.

Also, L2 will be in the opposite direction of L1; and I believe that the alternate statement for the cross product of a X b = absin(θ) accounts for this. Where a, and b are the magnitudes of the vectors of a and b.
 
One thing at the time. Let's try a few simple tings:
What is the magnitude of Labout the origin at t=0 ?
What is the magnitude of Labout the origin at t=v/(##\pi##R) ?

What is the magnitude of Labout P at t=0 ?
What is the magnitude of Labout P at t=v/(##\pi##R) ?

Do you have an expression for ##\vec r(t)## ?
 
Ah, I get the impression you have difficulty with
* distinguishing Labout (0,0)(t) from Labout (-R,0)(t)
* interpreting r x p

What do you mean with L1 and L2 ?
 
Hello Schulze

Your guess is incorrect .

Look at the figure attached .O is the origin.The coordinates of Q are ##R \hat{i}## whereas the coordinates of P are ##-R \hat{i}## .Let the particle be at M at any time 't' making an angle θ with the horizontal.

All you have to do is calculate ## m(\vec{PM} \times \vec{v})## .

1. First express θ in terms of v,R,t .
2. Write down the coordinates of M .
3. You already have coordinates of P given . So calculate ## \vec{PM}##
4. Find the component of velocity at point M in 'x' and 'y' direction . From that express velocity vectorially .

Now perform the cross product . Be careful with trigonometric identities .You will get the elusive 1 .
 

Attachments

  • circle.GIF
    circle.GIF
    2.5 KB · Views: 1,008
Last edited:
BvU said:
Ah, I get the impression you have difficulty with
* distinguishing Labout (0,0)(t) from Labout (-R,0)(t)
* interpreting r x p

What do you mean with L1 and L2 ?

That impression would be correct.

A friend of mine told me the right answer but wouldn't tell me how to arrive at this result. So that that is my next aim.

Answer: L = (mvR) cos((vt/R) + 1)
 
I did it folks!
 
Well, in that case we won't be needing the drawing I had already prepared when Tanya beat me to it :frown:.

For the sake of others listening in: what did you do and how did you do it ?
 
  • #10
\vec{L} = \vec{r} \times m\vec{v}

The angular displacement around the circle:
θ = ωt = \frac{vt}{R}

The vector from the center of the circle to the mass is then:
Rcos(θ)i + Rsin(θ)j (in the i and j directions)

The vector from the point P to the point of the mass is:
\vec{r} = R\hat{i} + Rcos(θ)\hat{i} + Rsin(θ)\hat{j}

\vec{r} = R[[1+cos(\frac{vt}{R})]\hat{i} + sin(\frac{vt}{R})\hat{j}

The velocity vector
\vec{v} = -vsin(\frac{vt}{R})\hat{i} + vcos(\frac{vt}{R})\hat{j}

So,
\vec{L} = \vec{r} \times m\vec{v}
becomes
\vec{L} = m v R {[1 + cos(ω t)]\hat{i} + sin(ω t)\hat{j}}
× [− sin(ω t)\hat{i} + cos(ω t)\hat{j}]

= m v R {[1 + cos(ω t)] [cos(ω t)] −[sin(ωt)] [− sin(ω t)] }\hat{k}

= m v R [cos(\frac{vt}{R})] + 1]\hat{k}
 
Back
Top