1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Totally confused acceleration and velocity of a particle in vectors(cross product)

  1. Nov 12, 2008 #1
    Totally confused!!! acceleration and velocity of a particle in vectors(cross product)

    1. The problem statement, all variables and given/known data
    The acceleration of a particle is given by a=vx(cxr) where r is the position, v is the velocity and c is a constant.

    Show that the following are constants
    (a)|v|
    (b)c.(rxv)
    (c)c.v-1/2|cxr|^2


    3. The attempt at a solution
    a=c(v.r)-r(v.c)

    d/dta=d/dt[c(v.r)-r(v.c)]
     
  2. jcsd
  3. Nov 12, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

    Something is constant if its time derivative is zero. |v| is constant if v.v is constant. Take d/dt(v.v). Remember d/dt(v)=a.
     
  4. Nov 12, 2008 #3
    Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

    Thanks for the help.

    Ive got part (a) and it equals zero. Its just im not too sure on what way to even start part (b) or (c). Should i use the cartesian components or how do i find out what r and c actually equal?
     
  5. Nov 12, 2008 #4
    Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

    (b) d(c.(rxv))/dt = c.(vxv) + c.(rxa) = c.(rxa)
    Try to show that this equals zero.

    (c) d(c.v)/dt = c.a
    d(1/2|cxr|^2)/dt = (cxr).(cxv)
    Try to show that these two equals each other.
     
  6. Nov 12, 2008 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

    How did you get a)? A very similar technique will get you b) and c). Start by taking d/dt of the expressions. For c) you might find the identity (axb).c=a.(bxc) useful.
     
  7. Nov 12, 2008 #6
    Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

    Thanks for both your help there, appreciate it alot.

    Just wondering though weejee, how did you get d(c.(rxv))/dt to be equal to c.(vxv)+c.(rxa)? where did the vxv and the rxa come from?
     
  8. Nov 12, 2008 #7
    Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

    Your welcome.

    The differentiation can act on either 'r' or 'v'. If it acts on 'r' it gives c.(vxv) and if it acts on 'v' it leads to c.(rxa).
     
  9. Nov 12, 2008 #8
    Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

    ahh, rite, i see what you mean now. Thanks alot.
     
  10. Nov 12, 2008 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

    It's called the 'product rule'.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Totally confused acceleration and velocity of a particle in vectors(cross product)
  1. Cross product vectors (Replies: 10)

Loading...