Polar Coordinates Homework: Converting to Cartesian and Strain Rate Tensor

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The discussion centers on converting a velocity field from polar coordinates to Cartesian coordinates and calculating the strain rate tensor. The velocity field is given as V = m/(2(pi)r) in the radial direction. The participant understands the conversion process but is uncertain about how to handle the angle theta in the context of the conversion. They express confusion regarding the relationship between r and the velocity field, specifically questioning if theta can be assumed to be zero. The conversation highlights the need for clarity in the conversion process and the correct formulation of the velocity field in Cartesian terms.
JSBeckton
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Homework Statement



The velocity field for a line source in polar coordinates (r,theta) is given by:

V=m/(2(pi)r) (in the "e" little r vector direction)

convert to cartesian and calculate the strain rate tensor.

Homework Equations



R=Sqrt(x2+y2);
Theta=ArcTan(Y/X);

Cartesian form:
X= R*cos(Theta)
Y= R*sin(Theta)


The Attempt at a Solution



I just need to convert this, i understand how to get the strain rate tensor. I know how to compute between polar and cartesian but need theta. Or am I suppposed to assume theta=0?

if theta=0 then y=0 andf x=r*sin(theta) but what do i use for r? If i solve for r i th efirst equation I get r=m/(2pi)

x=rcos(0)
x=m/(2pi)

This doesn't seem right becasue I need a velocity field in terms of V(u,v) where u and v are in terms of x and y.

Any help is greatly appreciated, Thanks
 
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I do not know what exactly you are asking, but if you are just asknig how to get from cartesian coordinates to polar ones than

theta=arctan(y/x) while R=sqrt(x^2+y^2)
 
I understand that as stated in my post under "Relevant Equations", thanks anyways
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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