An elementary treatment of the theory of spinning tops and gyroscopic motion, by Harold Crabtree.

ANSWERS TO EXAMPLES

137

12. Because a finite force cannot be produced instantaneously, and therefore some time must elapse before the maximum pull is attained.
At the first pull, a smaller force, which is increasing up to a
maximum, takes its share in producing momentum; but later this
inferior force is taken up in destroying the existing momentum
in the opposite direction to it, and by the time that it is again
producing momentum, the maximum force has been attained, and
thus the total work done in producing momentum=the maximum
pull x the length of the string. Thus the energy, and therefore the
momentum produced are greater in the second case.
13. The ratio of the energies produced-i.e. the work done in the two
cases, by pulling a string of length I with a force T, is
(T- T I T1
= e27rS - 1 e2vr.
14. (i) 1000 ft.-lb.-sec. (ii) 40400 ft.-poundals.
(iii) 16 rad./sec. (iv) 12~ seconds.
15. (i) 2182l- ft.-lb.-sec. units. (ii) 2304 ft.-poundals.
(iii) l1L seconds.
16. k2= j1f.2
17 550.400.121 550.605
17 49 = 11 H.P.
49 7 7
550.400.121 2750 44
' = 31a ft.-lbs.
49 7 7
18. (i) 96F ft.-poundals.
(ii) (a) 9-, X being the constant of proportion.
9k
The same forces are used in each case, but in inverse order, and
hence the result is the same.

(iii) 9X.

(iv) 31 X, = 2(-.
(iv) n+1 l V-WT= I

CHAPTER II.
PAGE 42.
2.,27i rad./sec.
5. */2 feet.

1. 1561 ft.-lbs.
4. 668 rad./sec.

3. 10 cm.

1. 43'816 ft./sec.
3. The radius= 42ag
11200. 7r
2.5. - 118i}7 ft.-lbs.
297

PAGE 45.
2. The wheels on the inside of the curve.
4. 1 2 R ft.-poundals.
6. -41 inches.