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58 MOTION OF A SPINNING BODY

Consequently, the total gyroscopic torque

L[= Kswswp + KS ws'wp'] = 0.94(75.6)0.29 + 0.11(314)0.29 = 30.6 slug-ft 2

The height X of the center of mass of the man and machine above the ground is, (29),

X = [170(36) + 300(16)]cos 0 = 23.2 cos 0 in. = 1.93 cos o ft. 170 +300

Substituting these values in the equation for tan 0,

30.6 slug-ft.2

tan B 1.93 cos 0 ft. = 0.03

= 4701b. wt. cos 0

Whence the inclination to the vertical due to gyroscopic effect is 0=sin'0.3=1°40'

Problem. Attached to the motorcycle of the preceding problem is a side car weighing 200 lb. and having the center of mass 30 in. from the cycle and 16 in. above the ground. Considering both centripetal and gyroscopic effects, find with what speed the machine must be moving around a curve of 100-ft. radius in order that the sidecar may be on the verge of rising off the ground.

Solution. When the motor and sidecar go around a curve, the system is acted upon by a torque tending to raise the sidecar. This torque is due to the resultant effect of the centripetal force pulling the car out of a straight path and the gyroscopic forces due to the change in direction of the spin-axes of the two flywheels and the three wheels that roll along the ground. The resultant torque tending to raise the sidecar is opposed by a torque due to the weight of the car. Each of these torques is equivalent to the product of a horizontal force along the radius of the circular path and the height of the center of mass of the system above the ground. The horizontal force applied at the center of mass of the entire moving system that is required to raise the wheel of the sidecar off the ground equals the sum of the centripetal force and the gyroscopic force due to the three wheels on the ground and the two flywheels. We shall now find the value of these four quantities and substitute them in the numerator of the second member of the equation,

The horizontal counteracting _ the force required to raise the sidecar force distance of the centroid above the ground

The torque, L [ = 30 2200] = 500lb:ft.

The distance X above the ground of the center of mass of the system consisting of man, motorcycle and sidecar is, (29),

X 1 _ 170 X 36 + 300 X016 + 200 X 16] = 21.1 in. = 1.76 ft. Whence, the horizontal counteracting force

Ft, L _ 500 lb.-ft. = 284 lb. wt.

X] 1.76 ft.

MOTION OF A SPINNING BODY 59 The centripetal force

0,2 670 v2 = 0.208 v2 lb. wt.

F` [ r ] 3

r 2.1(100)

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

where

Ks mk2] _ 325 14), [= 32.1(12 = 13.7 slug-ft 2

WS = 2 7x610300 = 136 rad. per sec.

and

WP [= r]= 220 = 1.1 rod. per sec.

Whence,

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.

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