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Angular motion of a particle

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m follows a path given by

    r = (Xo + at^2)x + bt^3 y + ct z

    where o, a, b and c are constants, t is the time. x, y , z should have hats on them showing they are position vectors (unit vectors).

    1. Find the angular momentum L of the particle about the origin.
    2. Find the force F required to produce this motion and verify explicitly that dL/dt = r x F

    2. Relevant equations

    3. The attempt at a solution

    L = r x p

    p = mv

    to get v differentiate r w.r.t time to get v = 2at.x + (3bt^2)y + cz

    Subsitute into the equation:

    L = ((Xo + at^2)x + bt^3y + ctz) x m(2at.x + (3bt^2)y + cz)

    Now to perform the cross product, im a bit unsure here. For the cross product i have this formula:

    r x p = x(ry.pz - rz.py) - y(rx.pz - rz.px) + z(rx.py - ry.px)

    where the x,y,z are unit vectors and the other x,y and z's are in subscript showing the particular component of each vector. So im a bit confused because do i just subsitute the constants from my equations of r and p or do i include their respective unit vecotrs associated with each component? But if i did that i would get for example x x y = z so would make quite a difference.

    Any thought guys, thanks
  2. jcsd
  3. Nov 18, 2007 #2


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    to be more clear, this is

    \vec{ r} \times \vec{ p} = (r_y p_z - r_z p_y) \hat{i} - (r_x p_z - r_z p_x) \hat{j} + (r_x p_y - r_y p_x) \hat{k} [/tex]
    In the equation above, r_x, p_y, etc are the components of your vectors [itex] \vec{r}, \vec{p} [/itex].
  4. Nov 18, 2007 #3
    Ok thankyou for explaining, i thought i was overthinking it. So i think that i get my value of L from r x p as:

    x(mc(Xo + at^2 - 3bt^3)) - y(mc(Xo - at^2)) + z(mb(Xo + t^2(a+3-2at^2))

    How do i find the Force required to produce this motion?

    Edit: do i just take the double derivative of r to get the acceleration and use F = ma. Then by taking the derivative of L i can show that my F multiplied by r should equal this derivative?
    Last edited: Nov 18, 2007
  5. Nov 18, 2007 #4


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    the force vector is simply the second derivative with respect to time of you rposition vector.
  6. Nov 18, 2007 #5
    I need to multiply it by m after differentiating it twice to get F right?
  7. Nov 18, 2007 #6
    Okay, im taking the secomd derivative of r, but a bit sure on my math. I have

    v = 2atx + 3bt^2 y + cz

    for a im getting it = 2ax + 6bt y

    but i wasn't sure if the z component completely disappeared or if the cz become just z - sorry im a bit rusty on this stuff. Anyone?
  8. Nov 18, 2007 #7


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    The z component of a is zero . (if you want, the derivative of [itex] c \hat{k} [/itex] is zero. )
  9. Nov 18, 2007 #8
    Thankyou, i had convinced myself of that also after a while. Just had to check.
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