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Re[2]: Free-body diagram Correction



First do the straight textbook problem of a car executing a banked
highway curve, showing that there is a preferred car speed which requires
no friction. As others have said, the only forces on the car are then
its weight and the normal force of support (which is inward and upward).

The resultant of these is the needed horizontal centripetal force. Then
simply observe that the banked highway curve is a section of a cone!



{{A different perspective. We observe the rotating cone from our
stationary position in the lab frame. Before the marble has been placed
on the cone, it has no angular momentum in our frame. We now place the
sphere on the cone's surface. For the sphere to begin spinning around
the symmetry axis of the cone, a torque must be exerted on the sphere
parallel to the symmetry axis, so that a nonzero angular momentum vector
develops parallel to the cone' symmetry axis. This requires exerting a
force on the sphere that lies in a plane perpendicular to the
cone's symmetry axis along a tangent to the surface of the cone.

The component of gravity that lies in the required plane points directly
at the axis of symmetry. *****OOOOOOPS. KEERECTION. Ignoring latitude
effects, gravity has no component in the aforementioned plane. Strike
that from the record. Which reminds me of an anecdote i just read.
David Niven, attending a gala, was speaking to an acquaintance at the
bottom of a long staircase when 2 women entered the staircase at the top.
Niven said 'that's the ugliest woman I've ever seen', to which his
acquaintance said 'that's my wife', to which David replied, 'I meant the
other woman', to which the acquaintance said 'That's my daughter'.
David, not missing a beat, 'I didn't say that'.*****


The same goes for the relevant component of the normal force exerted by
the cone's surface on the sphere. The only force that would have the
requisite component is friction, but the the cone's surface is
frictionless.

I conclude that no torque is exerted on the sphere that will cause it to
spin around the symmetry axis of the cone. Instead, the sphere will
slide down the surface of the cone as though it were on a frictionless
inclined plane.

Philip Zell
Zell@act.org}}

-Bob

Bob Sciamanda trebor@velocity.net
Dept of Physics
Edinboro Univ of PA http://www.velocity.net/~trebor
Edinboro, PA (814)838-7185

-----Original Message-----
From: Yvon Jean <phys@ntl.sympatico.ca>
To: phys-l@atlantis.uwf.edu <phys-l@atlantis.uwf.edu>
Date: Monday, April 20, 1998 9:15 PM
Subject: Re: Free-body diagram Correction


Sciamanda wrote:
It matters not whether the cone is rotating;
the object, however needs a specific velocity (equivalently,
rotational speed).

I tend to agree with him. The normal force will be the same in both
cases...but the students really have a hard time with this one. Please
enlighten!

If the inner surface of the cone is frictionless, then the sphere cannot be
dragged along by the cone's rotation.
The sphere has no angular momentum and therefore will exert no