Forces of Nature

Date: 10/9/95 at 13:38:56
From: Anonymous
Subject: sliding inclined rod (Physics)

Here is an elementary physics question which has been puzzling me.
I know virtually no physics, but I do have a good grasp of vector
mechanics, so if you could please phrase your answer in terms of
mathematical vector operations (instead of using physics terminology)
it would be much appreciated.

Okay, my understanding is that any motion in a system initially at
rest must be a result of Forces present in that system (and these
forces in turn can all be expressed as vectors).  Now, suppose we have
an "almost upright" rod which touches the (frictionless) ground at
the single point at its end and makes an angle Theta with the ground.
As far as I can see, the only Force involved would be Gravity, which
would be directed straight down.  (Though there may also be some kind
of "tension" Force, internal to the rod holding it together?  In any
case, these Forces would presumably be parallel to the rod and would
not affect my question).  

Now my intuition tells me that what will happen is that eventually the
rod will be flush on the ground and moving with a velocity parallel to
the ground.  Could you please explain all the forces which resolved
here?  I don't see how gravity alone (pointing downward) could result
in sideways motion.  Thanks.

Date: 10/9/95 at 14:11:49
From: Doctor Andrew
Subject: Re: sliding inclined rod (Physics)

There is one more force.  It is called the normal force and it pushes
against the rod from the ground in the vertical direction (perpendicular
to the ground).  It keeps the rod from going through the ground.  Its 
magnitude is equal and opposite to the vertical force the rod puts on 
it.  

You could construct a diagram which includes the rod, a vector 
representing gravity acting downward on the center of gravity of the 
rod, and a vector representing the normal force of the ground pushing on 
the end of the rod.  The y position of the bottom of the rod is fixed 
and there is no friction, so the x position of the center of gravity of 
the rod is also fixed.  The normal force exerts a torque on the rod 
causing it to rotate as it falls.  The magnitude of the normal force 
depends on the mass and moment of inertia of the rod. You can set up an 
equation for the y-displacement and acceleration, and another one for 
the rotation and the rotational acceleration.  The magnitude can be 
determined because the y-position of the center of the rod and the 
rotation of the rod are linked.  (y = l/2 sin a) where y is the 
displacement of the rod, l is the length of the rod and a is the angle 
of the rod with the ground.  The linear accelerations and velocities are 
tied with the rotational ones in the same way.

Hope this helps.

-Doctor Andrew,  The Geometry Forum