Understanding Polar Coordinate Unit Vectors

[ehild]

Yes, but does it move radially outward now?

ehild

 

[Panphobia]

I am not grasping what you are trying to ask...

 

[ehild]

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

 

[Panphobia]

it is 0 because dr/dt of a constant = 0

 

[ehild]

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

 

[Panphobia]

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?

 

[ehild]

Yes, the velocity vector is \vec v = 10 \hat e _{\theta} 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

 

[Panphobia]

I only needed the velocity vector in polar coordinate form. Thank you so much for the help!