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Angular momentum of time dependent particle motion

  1. Apr 8, 2014 #1
    1. The problem statement, all variables and given/known data
    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


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

    3. 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)
     
  2. jcsd
  3. Apr 8, 2014 #2

    BvU

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    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)
     
  4. Apr 8, 2014 #3
    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.
     
  5. Apr 8, 2014 #4

    BvU

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    One thing at the time. Lets 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)## ?
     
  6. Apr 8, 2014 #5

    BvU

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    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 ?
     
  7. Apr 8, 2014 #6
    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 .
     

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    Last edited: Apr 8, 2014
  8. Apr 8, 2014 #7
    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)
     
  9. Apr 9, 2014 #8
    I did it folks!
     
  10. Apr 9, 2014 #9

    BvU

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    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 ?
     
  11. Apr 9, 2014 #10
    [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]
     
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