Consider a disk under uniform circular motion. The acceleration of a particle on disk \(r\) units away from center is given by : \(a_c = \dfrac{v^2}{r}\) does that mean the acceleeration is inversely proportional to the radius ? why or why not ?

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Consider a disk under uniform circular motion. The acceleration of a particle on disk \(r\) units away from center is given by : \(a_c = \dfrac{v^2}{r}\) does that mean the acceleeration is inversely proportional to the radius ? why or why not ?

Mathematics
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If it's a rigid disk, then all particles have the same angular acceleration \(\alpha\), but their acceleration \(a = r \alpha\) so acceleration is actually directly proportional to radius here's the flaw with your argument: v is not constant. it also depends on the radius.
How do you explain it more lucidly to a 11th grader ?

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You can't say that time is inversely proportional to time just because \(t = \dfrac{t^2}{t}\) The top (numerator) of the fraction is a function of radius too so you can't say acceleration is inversely proportional to radius.
I could show him that \(a_c = \omega^2 r\), but how to really make him see that \(r\) is somehow more important than \(v\) ?
that is nice so what you saying is that just because something is there in the denominator wont make the quantity vary inversely
well you could either use that example about time the thing he'd need to understand is that the numerator is not constant, so you'd need to get something constant in the way. direct and inverse proportion always means that the rest is constant right?
yeah exactly
there was another question about circuits which @shrutipande9 asked in which she was confused about a similar thing: is power due to resistance \(\propto V^2\) or \(\propto V\) cuz \(P = VI = V^2/R\)
wow yeah same thing \[P=VI = \dfrac{V^2}{R} = I^2R\]
varying voltage does affect current but not resistance
who is depending on whom
yes! so it is proportional to \(V^2\) and since \(V \propto I\) we know \(I^2\) is also a choice

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