Question

quick clarification: NO MATH!

I feel like im getting confused can someone explain, when the radius value is decreased, how does the centripetal force change?

Answer 1

\[F_{Centripetal} = \frac{mv_{T}^2}{r}\]
Where Vt is the tangential velocity.

So, if m and v stay the same, but we decrease r by half, what happens to F?

Answer 2

it would get larger right?

Answer 3

I know you said no math, but this is the clearest way that I know to show you.

Let's say that we were a distance a from the pivot point. That is, r=a.

Our centripetal force would be:
\[F = \frac{mv^2}{a}\]
Easy enough, right? Let's see what happens when we decrease this distance by half.
r = (1/2)a
\[F = \frac{mv^2}{\frac{1}{2}a} = \frac{2mv^2}{a}\]

In general, we could summarize the behavior of F as we change r by just "ignoring" the things that aren't changing:
\[F \propto \frac{1}{r}\]or equivalently:
\[F \propto r^{-1}\]
where the "fish" sign means "is proportional to" or "varies as"
So F is proportional to one over r.

So now, if we want to see what will happen to F if we decrease (or increase) r by some factor, we just throw it in for r. Let's look at one third as an example. We want to decrease r to one third of it's current amount. What will happen to F?
\[F \propto (\frac{1}{3})^{-1} = 3\] So F will be increased by a factor of 3.

Answer 4

Make sense?

Answer 5

Makes sense, the Fc and radius are inversely proportional, so as radius decrease force increases.