Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: Newton's second law



This would seem ripe for an easy (secondary level) experimental test. Using
a motion probe (AKA sonic ranger) and a modified low friction dynamics car
(rigged with a silo full of shot and an aimable horizontal discharge
chute), display the usual motion curves (graphs) for position-time,
velocity-time and acceleration-time as the car is uniformly sent on its way
with the silo empty, then with the silo full but not discharging, then with
the silo discharging at right angles to the direction of motion, then with
the discharge aimed toward and finally away from the direction of motion.
The results should be interesting and easily related to the current discussion.

Tom Ford

At 04:12 PM 11/6/03 -0500, Bob at PC wrote:
Maddening, isn't it? :-)

Think of the difference between dropping stationary sand onto the moving
flatcar (basically the long discussion just concluded on this list about
rain falling into a flatcar) and having that same sand slowly leaking
out a hole at the bottom of the flatcar.

In one case you are accelerating the falling sand from a horizontal
speed of zero to a horizontal speed that is the same as the flatcar. You
are in essence creating a momentum, v*dm, for the falling sand each time
interval dt. If the flatcar is rolling freely and is not part of a
train, there are no external forces. The force creating the momentum of
the the fallen sand is counterbalanced by a force of the fallen sand
back on the flatcar. The term m*dv/dt represents the resulting
deceleration of the flatcar which has an instantaneous mass m (as
someone has already pointed out.) The sum of the two terms is zero.

In the second case, the sand leaks out, but keeps its horizontal speed
v. There is no horizontal creation or destruction of momentum for the dm
lost in each time interval, so the v*dm/dt term is not included when
evaluating dmv/dt.

The key here is that you are looking at the rate of change of momentum.
If dm has an associated momentum change then use the v*dm/dt term,
otherwise not.

Bob at PC

Justin Parke wrote:
>
> In a message dated 11/6/2003 3:08:09 PM Eastern Standard Time,
rlamont@POSTOFFICE.PROVIDENCE.EDU writes:
>
> > The v in
> > this case is the sideways speed of the sand, not the speed of the
> > flatcar down the track - which remains constant because
> > there is no
> > force in this direction.
> >
> > Bob at PC
>
> How do we know that "in this case" v is the sideways speed of the
sand? Sorry to sound so stupid...just trying to completely straighten
out my thinking on this.
>
> Justin Parke
> Oakland Mills High School
> Columbia, MD