Page 469 - Introduction to Continuum Mechanics

Principle of Moment of Momentum 453

Fig. 7.8

Solution. Let V c be a control volume that rotates with the sprinkler arms. The velocity of
water particles relative to the sprinkler is (Q /A)ei inside the right arm and (Q /A)(-ei) inside
the left arm. If p is density, then the total net outflux of moment of momentum about point
O is 2pQ(Q /A)&in0r 0^. The moment of momentum about O due to weight is zero. Since the
pressure in the water jet is the same as the atmospheric pressure, taken to be zero gage
pressure, there is no contribution due to surface force on the control volume. Now, since the
control volume is rotating with the sprinkler, therefore, we need to add those terms given in
Eq. (7.9.5) to the moment of forces. With x measured from 0, the first term of Eq. (7.9.5) is
zero, with & a constant, the second term of Eq. (7.9.5) is zero, with x=jte 1 and to - ^63, the
third term of Eq. (7.9.5) is zero. Thus, the only nonzero term is

which is the moment due to the Coriolis forces. Now, for the right arm, v = (Q /A)^
therefore,

and

Thus, the contribution from the fluid in the right arm to the integral in Eq. (i) is