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ganeshie8
 one year ago
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 ?
ganeshie8
 one year ago
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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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1If 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0How do you explain it more lucidly to a 11th grader ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1You 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I could show him that \(a_c = \omega^2 r\), but how to really make him see that \(r\) is somehow more important than \(v\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0that is nice so what you saying is that just because something is there in the denominator wont make the quantity vary inversely

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1well 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?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1there 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 \(P = VI = V^2/R\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0wow yeah same thing \[P=VI = \dfrac{V^2}{R} = I^2R\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1varying voltage does affect current but not resistance

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0who is depending on whom

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1yes! so it is proportional to \(V^2\) and since \(V \propto I\) we know \(I^2\) is also a choice
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