In uniform circular motion we know that $a_c=\frac {v^2}r=\omega^2r$.So,is $a_c$ directly or inversely proportional with $r$ and why not the other is true?
Thanks for any help.
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2 Answers
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In uniform circular motion, $v = v(r)$ so that $a_{c} \propto r$, not $a_{c} \propto r^{-1}$. In other words, $v$ is not a constant at all radii while $\omega$ is constant for all radii. So the second part of the expression is the one you want to look at in this regard.
Though the first half of the expression states that $a_{c}$ is only explicitly dependent upon $r^{-1}$, it is still implicitly proportional to $r$.
answered Jan 25, 2016 at 13:38
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In the first expression $v$ is the speed of a single point on the disk, actually any point that is a distance $r$ from the center. In the second expression, $\omega$ is the angular speed of the entire disk. Since $v$ and $\omega$ have very different definitions, the interpretation of those two equations is quite different.
answered Jan 25, 2016 at 13:29
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$\begingroup$ Exactly it is a simple mathematical reason. $\endgroup$ Commented Jan 25, 2016 at 13:36