Quantcast
Channel: Classical Physics
Viewing all articles
Browse latest Browse all 57941

Circular motion

$
0
0
I have seen the derivation of the centripetal acceleration formula a=v^2/r by saying r= rcosθi+isinθj=rcosωti+rsinωtj and differentiating twice. Since ω is constant we get a=-ω[itex]^{2}[/itex]r.

I've started looking at non-uniform circular motion where there is also the tangential acceleration vector component. I'm told that the component of acceleration directed towards the center of the circle has magnitude v^2/r, but I don't believe the original proof works because we assumed ω is constant, and now it isn't. Can this type of proof be made to work still?

Viewing all articles
Browse latest Browse all 57941

Trending Articles