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Motion of a particle given position vector.

 
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May6-12, 04:58 PM   #1
 

Motion of a particle given position vector.


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!
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May6-12, 05:40 PM   #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?
May6-12, 05:48 PM   #3
 
Quote by Steely Dan View Post
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?
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!
May6-12, 06:03 PM   #4
 

Motion of a particle given position vector.


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.
May6-12, 06:46 PM   #5
 
Is it supposed to look like a spring? And was I correct about the velocity?

Thank you for the help.
May6-12, 07:09 PM   #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.
May6-12, 07:10 PM   #7
 
Awesome! Thanks!
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