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Since F=mv^{2}/r. Does that mean the force required for centripetal acceleration is inversely proportional to the radius? If radius is more the lesser centripetal force is required, is this correct?
What's "v^{2}?"F=mv^{2}/r
That's if v is constant,radius is more the lesser centripetal force is required, is this correct?
Think again. What is "v?"v^{2}/r remains constant
That would be if F is a constant.F= mv2/r. So in this formula we know that as radius increases the speed or velocity increases
V is the velocity. But I didnt understand your point.Think again. What is "v?"
V is velocity which is distance/time. So you mean to say that the distance or the radius in this case is more so the velocity is more at the edge. So the Centripetal Force is more. Is this what you meant to tell me?What is "v" in terms of "r?"
mv2/r
inversely proportional to the radius?
I'm tryng to interpretate the law: in the first equation the force, that deviate the particle along the orbit, increases together with the speed if we fix the radius, and it decreases together with the radius if we fix the speed.Now there are 2 formulas for centripetal acceleration. 1. F= mv^{2}/r. So in this formula we know that as radius increases the speed or velocity increases. So the term v^{2}/r remains constant. So how can we prove from this formula that the Centripetal force is more if the radius is more.
But if we take the formula 2. F=mω^{2}r. we can easily prove that as the radius increases the Centripetal force required increases. But this inference cant be derived from the first formula.