Re: The old centrifugal force

[Vern Lindberg <vwlsps@RITVAX.ISC.RIT.EDU>]

On Friday, November 14, 2003, at 06:44 PM, Ludwik Kowalski wrote:

> The object is in the 5 o'clock position. I draw the
> m*g as a vertical vector pointing down. I decompose
> it into two components: radial (away from the center)
> and tangential (clock-wise direction). Then I draw
> the normal force vector

> which has the same length
> as the radial component of m*g but the opposite
> direction.

No. Repeat several times: "The normal force is the normal force. It
does not need to have any connection with the force of gravity. There
is no equation to predict the normal force in general. It is not the
component of the force of gravity. The normal force can only be
determined by using Newton's Second Law and information independently
determined about the acceleration. "

Skip circular and analyze a block on a horizontal surface (a) pulled by
a horizontal rope where N = mg (b) pulled by a rope angled above the
horizontal, N<mg (c) with the surface a floor of an elevator
accelerating up (or down), N>mg (or N<mg) and finally push the block
against a vertical wall with a push, P, at an angle theta above the
horizontal, N = P cos theta.

Now come back to your original problem.

> So far all is the same as when r is infinite
> (an object sliding along an inclined plane).
Dr. Vern Lindberg
585-475-2546
http://www.rit.edu/~vwlsps