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Re: The sign of g



At 13:59 -0500 9/3/02, Tina Fanetti wrote:

How do I explain to a student that when we choose up (+y) to be incre=
asing, we say that the sign of g is negative, regardless of whether t=
he ball is going up or going down. =20

My students seem to get that when the object is going up, g is negati=
ve. They want to make the ball going down positive g. But it stays =
negative because of the choice of axis. =20

I am at a loss because until I read Aarons I thought that it changed =
signs too...

I'll add my $0.02 worth to what everyone else is saying.

My guess is that the problem arises because the students are trying
to develop a problem-solving algorithm without trying to understand
physically what is going on. This is, of course, the reverse of what
should be happening. If they understand physically what is happening
, the algorithm for solution will follow naturally. Try to get them
to buy into this model: On the way up, the object is slowing down.
That means that something is pulling back on it--that means it is
pulling down. On the way down, the object is speeding up. That means
that something is pulling it ahead--that also means down. So whatever
it is that is acting on the object is acting in the same direction
during both halves of the trajectory. If we want to call the effect
of that action, its acceleration, then it follows that the
acceleration is pointed in the same direction throughout, namely,
down.

As has been said, g does not have a sign. It is the magnitude of a
vector which just happens to point in a downward direction. That
cannot be ambiguous, even if the coordinate system chosen happens to
be. Furthermore, g has nothing to do with the problem. g is something
that is there and can be measured whether the problem exists or not,
and it is *not* an acceleration. It happens to have units that are
equivalent to acceleration, but it is no more an acceleration itself
than a torque is an energy just because torque and energy share the
same units. In fact, an acceleration is only equivalent to the
magnitude of g in a very restricted set of circumstances: that is,
when an object is falling freely near the surface of the earth in
circumstances where the effect of air resistance on the fall can be
neglected.

In one-dimensional problems, it is not unreasonable to associate
direction with algebraic sign, so that, if one chooses up to be the
"positive direction" then down will be the negative direction. That
stays fixed throughout the entire problem. To change it in
mid-problem is to invite disaster. If we are doing problems where the
acceleration can be taken to be equal to the magnitude of g and
directed downward, then it must stay directed downward throughout the
problem, because g does not change direction for the convenience of
the problem. It is *always* directed downward. So if up is positive,
the value of the acceleration must always be equal to -g (assuming
that gravity is the only meaningful force in the problem, whether is
it posed that way or not). This can either be dealt with in equations
by writing -g, and using the magnitude of g, or by writing g and
always using g = -9.8 m/s^2. (All of this assumes that "up" is
invariably defined to be positive, which for beginning students, is
usually a good idea.)

If they resist this approach, ask them what the acceleration is at
the very top of the trajectory. I suspect some will argue vigorously
that it must be 0 (after all the object is at rest at the top, isn't
it?). This may have something to do with why they want g to change
direction between up and down legs of the problem. Dealing with that
issue may make the rest of it easier for them to swallow.

Acceleration is the name of a process (the changing of velocity) and
not the name of a direction. It is *always* defined in terms of delta
v/delta t, or dv/dt if one is teaching students who understand
derivatives. And delta v is further defined as v-final - v-initial
(similarly with delta t), in other words, changes are *always*
written as final value less the initial value, with (in one
dimension) the algebraic signs carefully carried with the numbers, so
that decreasing speed is acceleration and so is increasing speed.
There is no such thing as "deceleration." Acceleration can be
positive or negative depending on what the value of delta v is (delta
t is always positive). Beginning and ending velocities can be
positive or negative, and the above rule for acceleration will always
give the right answer.

For example, if the direction your car is facing is taken to be the
positive direction, and you are stopped, when you step on the
accelerator, your car will accelerate in the positive direction.
After you get up to speed, and step on the brakes, your car will
accelerate in the negative direction. From a stop, if you put the car
in reverse and step on the accelerator, the car will accelerate in a
negative direction. While going backwards, when you step on the
brake, the car will be accelerated in a positive direction.

The big conceptual mistake in their thinking, I believe, is that they
have associated acceleration with speeding up, so if the object is
slowing down, it must have negative acceleration, and if it is
speeding up, it must have positive acceleration. Get them to redefine
acceleration in their minds with changing velocity, which can be
positive or negative, whether the velocity itself is either positive
or negative.

This is not a trivial hang-up for the students and it is worth taking
some time to get them over the hurdle. It will pay off later.

Hugh
--

Hugh Haskell
<mailto:haskell@ncssm.edu>
<mailto:hhaskell@mindspring.com>

(919) 467-7610

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