Question

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 ?

Answer 1

@ParthKohli

Answer 2

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.

Answer 3

How do you explain it more lucidly to a 11th grader ?

Answer 4

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.

Answer 5

I could show him that \(a_c = \omega^2 r\), but how to really make him see that \(r\) is somehow more important than \(v\) ?

Answer 6

that is nice
so what you saying is that just because something is there in the denominator wont make the quantity vary inversely

Answer 7

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?

Answer 8

yeah exactly

Answer 9

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 \(P = VI = V^2/R\)

Answer 10

wow yeah same thing
\[P=VI = \dfrac{V^2}{R} = I^2R\]

Answer 11

varying voltage does affect current but not resistance

Answer 12

who is depending on whom

Answer 13

yes! so it is proportional to \(V^2\) and since \(V \propto I\) we know \(I^2\) is also a choice