Angular momentum of time dependent particle motion

[Schulze]

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(θ_r,p) = Rmv(sinθ_r,p)
θf = θi + (ω_it)
θf = θi + (vt/R)

The Attempt at a Solution

L₁ at position Q = R x p = Rpsin(θ_r,p) = Rmv(sinθ_r,p)

L₂ at position P = R x p
L₂ at position P = Rpsin(-θ_r,p)
L₂ at position P = Rmv(-sinθ_r,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)

 

[BvU]

1. Make a drawing
2. Realize that L = r X p is NOT Rmv(sinθ_r,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)

 

[Schulze]

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.

 

[BvU]

One thing at the time. Let's try a few simple tings:
What is the magnitude of L_{about the origin} at t=0 ?
What is the magnitude of L_{about the origin} at t=v/(##\pi##R) ?

What is the magnitude of L_{about P} at t=0 ?
What is the magnitude of L_{about P} at t=v/(##\pi##R) ?

Do you have an expression for ##\vec r(t)## ?

 

[BvU]

Ah, I get the impression you have difficulty with
* distinguishing L_{about (0,0)}(t) from L_{about (-R,0)}(t)
* interpreting r x p

What do you mean with L₁ and L₂ ?

 

[Tanya Sharma]

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 .

 

[Schulze]

Ah, I get the impression you have difficulty with
* distinguishing L_{about (0,0)}(t) from L_{about (-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)

 

[Schulze]

I did it folks!

 

[BvU]

Well, in that case we won't be needing the drawing I had already prepared when Tanya beat me to it .

For the sake of others listening in: what did you do and how did you do it ?

 

[Schulze]

[itex]\vec{L}[/itex] = [itex]\vec{r}[/itex] [itex]\times[/itex] m[itex]\vec{v}[/itex]

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

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:
[itex]\vec{r}[/itex] = R[itex]\hat{i}[/itex] + Rcos(θ)[itex]\hat{i}[/itex] + Rsin(θ)[itex]\hat{j}[/itex]

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

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

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

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

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