ing this equation, we obtain k (ln no)/n; (b) w= g — lovoi +TO.
1.94. F = (mv„ 2 1R) cost a.
1.95. F = —mco 2 r, where r is the radius vector of the particle
relative to the origin of coordinates;
F - m (1) V 2 x 2 + y^2.
1.96. (a) Ap = mgt; (b) I Ap I =
—2m(vog)Ig.
1.97. (a) p = at 3 /6; (b) s =
=--a-r 4 /12m.
1.98. s = (cot — sin cot) Fo/ma)^2 ,
see Fig. 6.
1.99. t = n/o); s = 2F 0 1m(0 3 ;
vmax = Foinuo.
1.100. (a) v=voe-trlm, 00;^ (b) v=vo —srlm,'
1.101. t=h (vo — v)(^) vov In (vo /u) '
1.102. s = — a^2 tan a, vmax = 17 (^1) a - sin a tan a
Instruction. To reduce the equation to the form which is convenient
to integrate, the acceleration must be represented as dv1dt and then
a change of variables made according to the formula dt = dxlv.
1.103. s a (t — to) 3 1m, where to = kmgla is the moment
of time at which the motion starts. At t < to the distance is s = 0.
1.104. v' = vo / kvVmg.
1.105. (a) v = (2F/mw) I sin (o)t/2)I; (b) As = 8Firnco^2 , (v)
= 4F/nnuo.
1.106. v = vo/(1 + cos cp). Instruction. Here w,^ —wx, and
therefore v = —vx^ const. From the initial condition it follows
that const = vo. Besides, vx = v cos q.
1.107. w = [1 — cos (11R)] Rgll.
1.108. (a) v = -1/2gfl /3; (b) cos % =
(^2) +11 ± 9112
3 (1+ 2 ) ' where 1=
= wo/g, 00 17°.
1.109. For n <1, including negative values.
1.110. When c0 2 R > g, there are two steady equilibrium posi-
tions: 0 1 = 0 and 0 2 = arccos (g1(0 2 R). When OR <g, there is
only one equilibrium position: 0 1 = 0. As long as there is only one
lower equilibrium position, it is steady. Whenever the second equi-
librium position appears (which is permanently steady) the lower
one becomes unsteady.
1.111. h z ((os 2 1v) sin cp = 7 cm, where 0) is the angular veloc-
ity of the Earth's rotation.
1.112. F = m g 2 w 4 r 2 ± (211(0)^2 = 8 N.
1.113. Feo r = 2m0) 2 r 1/1 (vohor)^2 ----- 2.8 N.
Off fit ad
Fig. 6.
Stotal