Question

Can someone explain how this equation is derived?

Answer 1

\[F_c=mr \omega^2\]
Relating certipetal force to angular velocity. Why isn't it
\[F_c= mr \alpha\]?

Answer 2

First off, we know from Newton's Law that
\[\huge F=ma\]Easy.
Additionally, we should know that centripetal acceleration is\[\huge \frac{v^2}{r}\]
Substituting, we get\[\huge F_c=m \frac{ v^2 }{ r }\]
Let's now refer back to arcs. We know that we can express arc length in terms of the angle and radius as such:\[\huge s=r \theta\]
Differentiating we get:
\[\huge v=r \omega\]
Plugging this into our equation from before, we now get
\[\huge F_c=m \frac{ v^2 }{ r }=m \frac{ (\omega r)^2 }{ r }=m \omega^2 r\]

Answer 3

@Dangazzm Do you get it now?

Answer 4

I understand everything except one small step. Why is centripetal acceleration related to V^2? wouldn't it be related like this. With linear acceleration?

\[a=r \alpha\]

Thanks in advance the explanation breakdown is EXACTLY what I needed.

Answer 5

For \(\large a_T=r \alpha\), a is the tangential acceleration and \(\alpha\) is the angular acceleration.

If you want the proofs (it's fairly long) I would suggest watching this video https://www.khanacademy.org/science/physics/two-dimensional-motion/centripetal-acceleration-tutoria/v/calculus-proof-of-centripetal-acceleration-formula
He goes into detail.
It's easier to just memorize the formula rather than derive it every time. Centripetal acceleration will always be the same for circular motion.

Here is more information about angular quantities which should prove useful!
http://spiff.rit.edu/classes/phys211/lectures/tang/tang_all.html

Answer 6

If you're not as advanced with calculus, then this might be a better video to watch https://www.khanacademy.org/science/physics/two-dimensional-motion/centripetal-acceleration-tutoria/v/visual-understanding-of-centripetal-acceleration-formula

Answer 7

That was genius, thanks for finding that wonderful khan academy video I got some things to mull around in my head but I get it now!

Thanks again!

Answer 8

You're very welcome! Glad I could help X)