1. Limited time only! Sign up for a free 30min personal 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!

Motion of a particle given position vector.

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data

    A position vector of a particle at a time t is r=icost +jsint +kt ; show the speed and the magnitude of the acceleration is constant. Describe the motion.

    2. Relevant equations

    v = dr/dt
    a = dv/dt

    3. The attempt at a solution

    Could someone let me know if I am doing this correctly?

    I derived the position to find the velocity:

    v = dr/dt = -isint +jcost + 1k

    Then derived the velocity :

    a = dv/dt = -icost -jsint

    Then found the magnitude of the acceleration:

    mag(a) = sqrt ( cos^2(t) + sin^2(t)) = 1 , which is constant.

    Motion- increasing oscillation? How would I show this?

    Thanks!
     
  2. jcsd
  3. May 6, 2012 #2
    Don't forget to find the speed (magnitude of the velocity).

    Think about just the two-dimensional [itex]x,y[/itex] motion. What kind of motion would that be? Then notice that the [itex]z[/itex] component just linearly increases in one direction. What kind of shape will be created this way? And how will your particle move along that three dimensional shape?
     
  4. May 6, 2012 #3
    I took the magnitude of the velocity and got sqrt( sin^2 + cos^2 +1) so sqrt(2) , so it would also be constant.
    Would the particle just be moving around a circle? How could I prove this?

    Thanks!
     
  5. May 6, 2012 #4
    Well, yes for the two-dimensional case it would be circular motion. If you don't recognize the form, try picking various values of [itex]t[/itex] and plotting them on a two dimensional graph to see it.
     
  6. May 6, 2012 #5
    Is it supposed to look like a spring? And was I correct about the velocity?

    Thank you for the help.
     
  7. May 6, 2012 #6
    Exactly, it's the shape of a helix. The motion is circular, but it's traveling upwards along the surface of a cylinder with time.

    And yes, you were right about the speed.
     
  8. May 6, 2012 #7
    Awesome! Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook