🚀 go-pugleaf

On Tue, 16 May 1995, David J. Abineri wrote:

> As the one who posed the question "what happens to a rigid meter stick when a
> force is momentarily applied perpendicular to the stick at one end?", I have
> read all of the responses with a great deal of interest.
> I find myself, however, not much closer to the answer (which I suspect is that
> the force causes both linear and rotational accelerations).

Well, none of the messages I read gave the entire answer, completely.
They merely pointed in the direction of the answer. No satisfactory
answer can be given without calculus, IMO.

> In my high school classroom I dutifully draw the force diagram and dutifully
> quote F=ma.

Fine. There's but one force acting on the stick, so it produces an
acceleration of the center of mass, as given by F = ma. But why an
acceleration measured at the center of mass and not some other point? If
you say this, you must be prepared to establish or prove it
mathematically.

> There is a net force so there must be a net acceleration. But, how do I
> explain the rotation?

> I suspect that the center of mass resists a change in
> its motion or, as someone said, "there is an 'inertial' force but how do I
> explain the existence of such a force.

I also have problems with this language, for it is an assertion which
hasn't been previously established or proven in the course. I don't have
the foggiest idea what an 'inertial' force is in this context. It smacks
of something invented for a pedagogical purpose, to avoid doing the
problem correctly.

> If indeed there is rotation, then one must ask how much rotational
> acceleration occurs and how much linear acceleration occurs? I don't know how
> to answer that one either.

The answer *was* given: Torque = (moment of inertia)x(angular velocity).
For this equation, torque and moment of inertia and angular velocity must
be expressed about the same common axis, through the CM.

> I firmly believe that there is no couple involved in this problem. But there
> does seem to be a torque applied by the single force.

No, I see no couple here. Just a single force, whose torque about the
center of mass is responsible for the rotation. When the sum of all
torques is zero, one may take the center of torques wherever one pleases,
and you still get zero net torque. But when the sum of torques is *not*
zero (as it is here), it matters very much where you take the center.

> I hope I have made my question clear and that we can continue to pursue it.
> Thanks to everyone who contributed. David Abineri    dabineri@tso.cin.ix.net

Quite clear. The above comments, most made already by several folk here,
are the way to a numeric answer. But do we *understand* what underlies
the procedures? Do we (and the students) grasp the concepts and the details?

While I was composing this message, a message from John Mallinckrodt
appeared, answering your question quite nicely, along these lines.

However, I suspect there are some underlying questions students still have
about the validity of the procedures used. How does the stick 'know' where
its center of mass is? Does it matter whether the body is rigid?

Consider if the body (the stick) were not rigid, but were, say a straight
string of beads, or a chain, struck by a force at one end. The outcome
would be quite different, wouldn't it? Suppose the 'stick' were just a row
of blocks or dice. Striking one end would propel one of the blocks and not
disturb the others. These observations convince us, and the students, that
the *rigidity* of the body is important to the outcome. And what is
rigidity anyway? It is due to internal forces within the body, the forces
which hold it together and tend to preserve its shape. That's how the rest
of the body 'knows' it has been hit somewhere (at one end), and how the
external force produces a change throughout the body in the size and
direction of these internal forces (there's a momentary deformation of the
body). Someone alluded to that here when they spoke of a 'sound wave'
propagating down the stick.

All of this is the stuff of the section titled "Rigid Body Motion" in
nearly any introductory college Mechanics text. Warning: it is bound to
use calculus.

There's an underlying important theorem which states that the motion of a
rigid body can be decomposed into two motions: "Translation of the CM"
and "Rotation about the CM". The kinetic energy of the body consists of
the sum of the individual mv^2/2 kinetic energies of each of its parts (a
calculus concept). This can be decomposed into two parts: (1) MV^2/2 where
M is the total mass of the body, and V is the translational velocity of
its center of mass. (2) IW^2/2, where I is the moment of inertia about
the CM and W is the angular velocity about the CM. Thus the conservation
of energy equation can break the kinetic energy into these two terms.

But now consider momentum. There's linear momentum, MV, which is the
body's total mass times the velocity of its center of mass. This can be
changed by an impulse, Fx(delta)t. There's also a *different kind* of
momentum, *angular momentum* which is IW. It can be changed by an angular
impulse, (torque)x(delta)t. (delta) means 'change in'. So to calculate the
outcome of problems of this kind requires *three* equations: energy
conservation, linear momentum conservation, and angular momentum
conservation.

I've tried to avoid *explicit* calculus, but at rock bottom these kinds of
problem *are* calculus problems, if you are going to include all steps and
concepts without unproven assertions and hand-waving. You need the
calculus to prove the theorems you are using. You will notice that the
rigidity of the body does not enter into John's solutions explicitly, but
it does enter into the theorems he uses.

        -- Donald