- #1

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I can see that If ##\vec r## unit vector is,##\vec r=(cosθ,sinθ)## then ,

##\frac {d\vec r} {dt}=\vec θ##

Like calculating ##\vec v=\frac {d\vec r} {dt}=R.\frac {dθ} {dt}.\vec θ##

I don't understand why we use this ? Just to make things easier ?

2-And we know that ##\vec a=R∝\vec θ+Rw^2(-\vec r)##

here

##\vec {a_t}=R∝\vec θ## and ##\vec {a_r}=Rw^2(-\vec r)##

I know that If ##a_t## is zero then it becomes uniform circular motion.But In main equations there is an non-zero angular acceleration.So it means do we have to exert ##\vec F=m(-R∝\vec θ)## to create a uniform circular motion ?

Note:This is not a homework question so,I didnt ask there