The following is intended only for the pedantically deprived.
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The side thread on units and dimensions scratches a minor itch of mine
about rotational dimensions and, particularly, the peculiar gyrations (so
to speak) that we go through to accommodate the radian. I'm going to
argue that the dimensions of torque are NOT (or at least SHOULD not be)
the same as those of energy.
All equations in physics are tyrannical with respect to dimensional
consistency. With only a few exceptions, however, they are completely
neutral about units. ANY units carrying appropriate dimensions are
perfectly acceptable. For instance, in the equation
x = x_o + v_o * t + 1/2 a * t^2
we can specify x_o in furlongs, v_o in parsecs/millisec, t in weeks, and
a in inches/(hour fortnight) and get a perfectly correct (if unwieldy)
answer like
x = 73.2 furlong + 1.65 x 10^-16 parsec week/millisec
+ 3.65 x 10^6 inch week^2/(hour fortnight)
A few unit conversions would be VERY nice at this point but are NOT
necessary.
Such is NOT (or at least APPEARS not to be) the case when it comes to
rotational equations. We tell students that some equations like KE = 1/2
I w^2 ONLY work if we express w in rad/sec or at least convert all
angular units to radians (and then mysteriously throw them out!) before
presenting the final answer.
It needn't be this way. Suppose we take seriously the idea of an angular
dimension (call it "A") and a quantity r' which is the tangential
distance per unit angle at some specific distance from an origin (to be
carefully distinguished from the quantity r which would still be the
distance from that origin.) Then r' would have dimensions L/A and, for a
point 1.00 meter away from the origin, we would have
r' = 1.00 m/rad = .0175 m/degree = 6.28 m/cycle
One can now go back through all of rotational kinematics and dynamics
rewriting equations using r' wherever appropriate and get things like the
following
QUANTITY "FORMULA" DIMENSIONS
angular momentum r' m v M L^2/(T A)
torque r' F M L^2/(T^2 A)
rotational inertia r'^2 m M L^2/A^2
angle r/r' A
Now you can calculate, for instance, the rotational kinetic energy using
the usual equation KE = 1/2 I w^2 with no need to use radians. For
instance if I is specified in kg m^2/deg^2 and the angular velocity is
specified in cycles/sec we would get a perfectly acceptable energy
specified in Joule (cycle/degree)^2. Of course one may now WANT to
convert that quantity to Joules using the simple conversion factor
cycle/degree = 360, but it is NOT necessary. And the radian is no longer
any more special or mysterious than any other unit.
BTW, lest anybody get me wrong, I'm not quite ready to advocate actually
using this system for the introductory course, but I do think it has some
significant advantages in terms of consistency.
John
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A. John Mallinckrodt email: mallinckrodt@csupomona.edu
Professor of Physics voice: 909-869-4054
Cal Poly Pomona fax: 909-869-5090
Pomona, CA 91768 office: Building 8, Room 223
web: http://www.sci.csupomona.edu/~mallinckrodt/