Mathematics of Physics and Engineering

Rigid Bodies 83

EXERCISE 2.2.16.C Verify that the vectors i, j, k define a principal axis
frame. Hint: Iy = Iz = Iz = 0, as seen from (2.2.41) and the symmetry of
the rod.

The motion of the rod is a 2-D rotation around the pin at O, and the

vector k* is not rotating. There are no internal forces in the rod to affect

the motion. As a result, the angular velocity vector is UJ = ui*z k* —Ok*,

and the Euler equations (2.2.35) simplify to I*z —^ = T^MI, or

M{P+a?).. (B)

^ V~^1 CMz' (2.2.44)

where TQM'Z is the rc*-component of the external torque around the center

of mass.

To simplify the analysis, we ignore air resistance. Then the external

torque T*jM' around the center of mass is produced by two forces: the force

of gravity W and the force Fpin exerted by the pin at O. Since the rod

is uniform, the torque due to gravity is zero around the CM. To compute

Fpin we now assume that the frame fixed at O is inertial and Newton's

Second Law (2.2.5) applies. The external force on mass M at the CM is the

sum of the total gravity force, W = Mgi, and the reaction of the pin FPin.

Hence, Fpin = M rcM — W. Since CM moves around O in a circle of radius

£/2 we use (1.3.27) on page 36 to obtain the components ar, ag of TCM

in polar coordinates with origin at O: ar = {—1/2)UJ^2 r, a$ = (^/2)w 6.

The point of application of Fpin relative to CM has the position vector

-rCM = -{i/2)r. Then

TCM = ~rCM x Fpin + 0xW = --rx Fpin.

Hence,

T*C(M =4?X {MV,CM -W) = ^(^Lbrx0-gvxi^

-IM (I.. „, .»

2 V2 -uj + gsmv I K.

Substituting in (2.2.44, we get (M(£^2 + a^2 )/12)6 = -(£M/2)({£/2)u> +

g sin 6) or {(£^2 + a^2 )/l2)0 + (£^2 w/4) + (^sin0/2) = 0, or, with w = 0,

({At^2 + o?)/l2)e = -{gl/2) sinS. (2.2.45)