MOTION OF A SPINNING BODY 59 The centripetal force
0,2 670 v2 = 0.208 v2 lb. wt.
F` [ r ] 3
The gyroscopic force due to the three wheels rolling along the ground, 3(15)12 v v
Kswswp = 32.1 1.165 100 = 0.0068 v2lb1 wt. Fg [ = X ] 1.76
The gyroscopic force due to the two flywheels
0.11 2 ~r 3000 v v
Ks'ws'wp' 60 X 88 100 = 0.0022 v1 lb. wt.
Fg [ - X ] 1.76
Substituting these values of Fl,, Fe, Fg and Fg' in the equation stated in words at the beginning of this solution,
284 = (0.208 + 0.0068 + 0.0022)v2
Whence, the speed with which the machine must move around the given curve in order that the sidecar may be on the verge of rising off the ground is
v= 36.15 ft. per sec. = 24.6 mi. per hr.
Problem. A rotary airplane engine of 325 lb., and radius of gyration 14 in., makes 1300 r.p.m. when the plane has a speed of 150 mi. per hr. Find: (a) the gyroscopic torque exerted on the plane when making a turn of 200ft. radius; (b) the force that must be applied at the tail when the distance between the center of pressure on the tail and the center of mass of the airplane is 15 ft.; (c) the mass that a similar engine of radius of gyration 14 in., would need to have in order that under similar conditions the pilot could produce a force on the tail amounting to 150 lb. wt.
Solution. (a) The torque is about a transverse axis and has a value
L = Kswswp
Ks mk2] _ 325 14), [= 32.1(12 = 13.7 slug-ft 2
WS = 2 7x610300 = 136 rad. per sec.
WP [= r]= 220 = 1.1 rod. per sec.
L [ = Kswswp] = 13.7(136)1.1 = 2050 lb.-ft.,
(b) The force on the tail
F [ = X] _ 2050 lb., ft. = 136.5 lb. wt.