Motion in Polar Coordinates problem

AI Thread Summary
The discussion centers on a problem involving a particle described by the polar coordinate equation r=be^[Zcot(a)], where Z represents the angle theta. The goal is to demonstrate that the velocity and acceleration vectors form angles a and 2a with respect to the origin O. A participant initially struggles with the calculations but eventually finds success by simplifying the velocity expression and utilizing the dot product to determine the angles between vectors. The conversation concludes with a resolution, highlighting the collaborative nature of problem-solving in mathematics.
Master J
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Been looking over past exam questions and came across this one. Its in polar coordinates:

A particle P describes the curve r=be^[Zcot(a)].

Show that the velocity and acceleration vectors have angles a and 2a with OP (O is the origin).

Z is actually theta, the angle the position vector makes with the x axis. Its derivative to time is w, a constant here. B and a are constants.


I'm stumped!


I work out v as: bcot(a)we^(Zcot(a)).[e_1] + bwe^('').[e_2]

Even with them worked out I am stumped as to show that^^^^

Any pointers?

Cheers guys!
 
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Hi Master J,

Master J said:
Been looking over past exam questions and came across this one. Its in polar coordinates:

A particle P describes the curve r=be^[Zcot(a)].

Show that the velocity and acceleration vectors have angles a and 2a with OP (O is the origin).

Z is actually theta, the angle the position vector makes with the x axis. Its derivative to time is w, a constant here. B and a are constants.


I'm stumped!


I work out v as: bcot(a)we^(Zcot(a)).[e_1] + bwe^('').[e_2]

I would suggest first simplifying this expression for v, using the expression for r given in the problem. Then, since they ask for the angle between two vectors, that suggests using the dot product (since one definition of the dot product explicitly includes the angle between the vectors). Do you get the answer?
 
SOlved! Thanks for the help.
 
Sure, glad to help!
 
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