Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

194 Aptitude Test Problems in Physics

spring is F2 = 2Ften sin a (see. Fig. 48). Accord..

ing to Hooke's law, F2 = (1.51 — 21 sin a) k ,

where k is the rigidity of the spring. As a result ,

FI = 1.51k cot a — 21k cos a.

In order to determine the period of small oscil-

lations, we must determine the force AF acting

on the load for a small change Ah in the height of

the load relative to the equilibrium position hi) =

21 cos a 0. We obtain

AF (^) 1.51k A (cot a) — 21k A (cos a),
where
A (cot a) —
d:cot a
Aa
Aa
da /c4=- a„ sine a 0
A (cos a) — sin ao Aa.
Consequently, since Ah = —2/ sin ao Aa, we find
that
AF, —1.5k
/ Au
+2k1 silt ao Aa
sine at,
— 5k1 Aa = — 5k Ah
because sin ac, = 1/2.
The period of small oscillations of the load can
be found from the formula T = 2n ir WIWO, where
in is the mass of the load determined from the
equilibrium condition:
1.5k/ cot ao — 2k1 cos ao = ing,
in =
kl/g
Thus,
T 1/r "
lOg
1.97. At each instant of time, the kinetic energy of
the hoop is the sum of the kinetic energy of the
centre of mass of the hoop and the kinetic energy