Motion of a particle given position vector.

[tcanman]

Homework Statement

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.

Homework Equations

v = dr/dt
a = dv/dt

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!

 

[Steely Dan]

Don't forget to find the speed (magnitude of the velocity).

Think about just the two-dimensional $x,y$ motion. What kind of motion would that be? Then notice that the $z$ 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?

 

[tcanman]

Don't forget to find the speed (magnitude of the velocity).

Think about just the two-dimensional $x,y$ motion. What kind of motion would that be? Then notice that the $z$ 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!

 

[Steely Dan]

Well, yes for the two-dimensional case it would be circular motion. If you don't recognize the form, try picking various values of $t$ and plotting them on a two dimensional graph to see it.

 

[tcanman]

Is it supposed to look like a spring? And was I correct about the velocity?

Thank you for the help.

 

[Steely Dan]

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.

 

[tcanman]

Awesome! Thanks!