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

ROTATION ABOUT A FIXED AXIS

27

if 0 be the angle through which G moves in destroying the kinetic
energy,
332. 0= 187.
0 375.3 radians
radians
57r
75.3 1
= -.- revolutions
r 27r
= 5 (taking r2 = 10)
=111 as before.
4. If the mass of the top in example (2) be 8 oz., its radius of gyration
about its axis of symmetry 2 ins., and the pulling force be 10 lbs. weight
throughout, how many revolutions a second will it be making when all the
string is unwound?
The work done=10.3.32 ft.-pdls.
If o=the angular velocity generated, in radians per sec.,
1
kinetic energy =- 0o2
2 2 6O
Equating the K.E. to the work done, we get
(02=10.3.32.4.36;.'. 0=96115;
484115.'. number of revolutions per sec.=48 -Notice that in this case the size of the spindle does not affect the problem.
If the spindle were larger the string would be unwound in fewer turns, but
in a shorter time. The force would act at a longer arm, and the power or
rate of work would be greater, but the actual work done, or kinetic energy
generated, would be the same. For the same reason the string need not be
" thin."
5. To a wheel and axle of mass 48 lbs. and radius of gyration 6 ins. is
attached a weight of 13 lbs. by means of a rope wound round the axle.
Taking the radius of the latter as 4 ins., and neglecting the weight of the
rope, find the velocity of the 13 lbs. weight after it has descended 52 ft.
We shall equate the work done by gravity on the whole system to the
kinetic energy of the system. Let v ft./sec. be the required velocity of
the weight: then the corresponding angular velocity of the wheel is
v -=3v rad./sec.
The K.E. of the whole system
1 62 1
-. 48. 22(3v)2 13 v2
121 ft.-pdls.
Work done by gravity on the system
Work done by gravity on the system.-pd
= 13. 32.52 ft.-pdls.