🚀 go-pugleaf

On Mon, 15 May 1995, Guntbert Reiter wrote:

> John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU> wrote:
>
> >Mervin Koehlinger writes:
>
> >>Forces produce translation; couples produce rotation.
>
> I prefer: Forces cause acceleration (not necessarily movement)
> while couples cause a torque (not a rotation).

You've still not gotten it right:

Net force produces acceleration (F = ma), while net torque produces
angular acceleration: (T = I*alpha) where T is net torque and 'alpha' is
angular acceleration.

A couple is defined to be the torque produced by two *equal* size and
oppositely directed forces with a displacement between their lines of
action. Any *good* freshman college physics book, or an introductory
engineering mechanics book will confirm these standard definitions and
laws. The torque due to a couple is obviously Fd where F is the size of
either one of the forces and d is the perpendicular displacement between
them.

> >Sorry, that does not compute.  A rod in "free space" will be set into
> >rotation by a SINGLE force as long as that force is not applied along
> >a line that passes through the rod's center of mass.
>
> John: It DOES compute. Imagine the impact resulted in pure
> movement without rotation. More exactly spoken: the result would
> be a linear acceleration of the CM. By Newton's laws (inertia...)
> we know that from this acceleration the CM should "feel" (I just
> can't remember the correct term) a corresponding force pointing
> in the opposite direction. So these two forces build a couple and
> produce a torque.

The CM (a mathematically computed point) can't "feel" anything. And
Newton's laws say nothing directly about the CM. Once you have *defined*
the CM you can use Newton's laws to derive things about it, but please
use them correctly.

I *assume* (though it's not at all clear from your paragraph above) that
the 'corresponding force in the opposite direction' refers to the reaction
force in Newton's third law. But, no, it can't, for the reaction force
described in that law acts on a *different* body. Newton's third, properly
stated is: "If body A exerts a force F on body B, then body B exerts an
equal and opposite force (-F) on body A." In the problem posed, a force
applied to a stick in space, reaction force does not *in any way* enter
into the solution of the problem (determining the subsequent motion of
the stick). That motion will be a combination of translation and
rotation, a result of the applied force, and the torque associated with
that applied force. One force *only* is involved in this problem, as
the problem was originally posed here.

If, and only if, that force were applied so that its line of action passed
through the CM of the stick, then the stick would move in translation only,
with no rotation. Refer to any introductory engineering mechanics book,
particularly the chapters on motion of rigid bodies. A recent book I find
on my shelf which has good treatments of this sort of problem is _An
Introduction to Dynamics, 3rd ed._ by David J. McGill and Wilton W. King.
PWS-Kent Publishing Company, 1995. The relevant equations are Euler's
First and Second Laws:

         __
         \  F  = ma
         /_  i     cm
        all i
                                            /
         __      __                 __      |
         \  M  = \  r Xm a    or    \  M  = | rXa dm
         /_  P   /_  i  i i         /_  P   |
                                            /


For a continuous mass distribution you'd use the second (integral) form.
P is 'any point', M-sub-P designates moments about P, the r-sub-i are
position vectors of the individual point masses m-sub-i, and a-sub-i are
their accelerations. X is cross product, and M, r, and a are vectors. The
location of the center of mass of the system of masses is *defined* by
              __
              \ m R                       /
              /_ i i                    1 |
        r   = ______       or,    r   = _ | R dm
         oC      m                 oC   m |
                                         /

where r-sub-oC is the vector from point o in a fixed, inertial frame to the
point C which is the center of mass. m-sub-i is one of the point masses,
R-sub-i is the vector from o to m-sub-i. Again, the integral form is for
a continuous distribution of mass (rigid body).

Anyone who attempts to do problems like this without having had a
junior-level mechanics course using calculus is asking for trouble. You
doubt me? Then try this problem:

During the early days of the space program, one method for reducing the
spin of a space probe after launching was to unfurl a couple of antenna
wires with weights on their ends, which were initially wrapped around the
shell of the cylindrical instrument package.

Imagine a cylindrical satellite of moment of inertia I and radius R1
spinning with angular velocity W1. Two wires are wrapped around it, with
two weights m. These are released, and unfurl so that the weights are now
extended to distances R2 from the center of the satellite. Calculate the
new (smaller) value of angular velocity, W2.

Do this by conservation of angular momentum:

(I + 2mR1)W1 = (I + 2mR2)W2    etc...

Do it by conservation of angular kinetic energy:

(I + 2mR1)(W1^2)/2 = (I + 2mR2)(W2^2)/2    etc...

These do not give the same answer. Why? Which is wrong? Why?


        -- Donald