• Support PF! Buy your school textbooks, materials and every day products Here!

Polar Coordinate Unit Vectors

  • Thread starter Panphobia
  • Start date
  • #26
435
13
so when you are doing the derivative of the position vector in polar form, do you have to use product rule to differentiate, because you need to differentiate the unit vectors too.
 
  • #27
ehild
Homework Helper
15,492
1,874
so when you are doing the derivative of the position vector in polar form, do you have to use product rule to differentiate, because you need to differentiate the unit vectors too.
Yes.

ehild
 
  • #28
435
13
so velocity = (dr/dt)*Er + 5*(dθ/dt)*Eθ?
 
  • #29
ehild
Homework Helper
15,492
1,874
V= (dr/dt)*Er + r*dEr/dt. And dEr/dt=dθ/dt Eθ.

What is dr/dt equal to in the problem?


ehild
 
  • #30
435
13
It is how fast it is moving radially outwards in the direction of Er. Right?
 
  • #31
ehild
Homework Helper
15,492
1,874
Yes, but does it move radially outward now?

ehild
 
  • #32
435
13
I am not grasping what you are trying to ask...
 
  • #33
ehild
Homework Helper
15,492
1,874
The particle moves along a circle of radius R=5. You want its velocity vector in terms of Er and Eθ. Er is multiplied by dr/dt. r is the distance of the particle from the centre. Does that distance change while the particle moves along a circle? So what is the value of dr/dt?

ehild
 
  • #34
435
13
it is 0 because dr/dt of a constant = 0
 
  • #35
ehild
Homework Helper
15,492
1,874
dr/dt is not 5. It is zero, as the distance from the origin does not change. So how can you write the velocity vector? And what about the acceleration?

ehild
 
  • #36
435
13
Yea after I wrote the wrong answer, I thought about it and changed it. But the velocity would just be this right? v = 5*(dθ/dt)*Eθ and the acceleration would be 0?
 
  • #37
ehild
Homework Helper
15,492
1,874
Yes, the velocity vector is [tex]\vec v = 10 \hat e _{\theta} [/tex] in this case.

The acceleration is the time derivative of the velocity. It is not zero, as the velocity changes direction. You need the derivative of Eθ now. Go back to #20 and figure out how is it related to Er.

ehild
 
  • #38
435
13
I only needed the velocity vector in polar coordinate form. Thank you so much for the help!
 
Last edited:

Related Threads on Polar Coordinate Unit Vectors

  • Last Post
Replies
1
Views
2K
Replies
1
Views
4K
Replies
9
Views
843
Replies
1
Views
1K
Replies
8
Views
2K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
16
Views
191
Replies
9
Views
7K
  • Last Post
Replies
7
Views
4K
Top