1) The "stationary table" frame is presumably an inertial frame - there is
no fictitious (-ma) force in this frame. F=ma simply becomes kmg=ma.
2) A centripetal force is a force applied at right angles to an object's
velocity. This has nothing to do with its "conservativenes". A satellite
in circular orbit experiences a conservative, gravitational, centripetal
force. An object adhering to a spinning turntable by means of friction
experiences a non-conservative centripetal force.
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net http://www.velocity.net/~trebor
----- Original Message -----
From: "Muhsin Ogretme" <ogretmem@BOUN.EDU.TR>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, May 06, 2003 4:16 PM
Subject: A free b.d. and a centripedal force question
| 1. Assume there is a block of mass M at rest on a frictionless table.
| There is another block of mass m at rest on top of M. But there is
| enough coefficient of friction between m and M, so that when M starts
| to accelerate with "a" under the action fo a force F in +x direction,
| m does not slip but they move together. When I consider the forces
| acting on m (Free body diagram) in x-axis, i can identify the
| friction force acting in +x direction which equals normal force
| (equal to mg) times the coefficient of friction k, and the fictious
| force (ghost force) in -x direction which is equal to ma. How would
| you write newtons second law for m: kmg-ma=ma ? which means kmg=2ma ?
| You reference frame is stationary table. I am almost sure about this
| analysis. I just wanted to check it with you.
| 2. Is the centripedal force in a uniform circular motion a
| nonconservative force? I believe it is neither cons. nor noncons.
| Your ideas?
| Cheers,
| Muhsin Ogretme