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1.70. '1 V-21/(3W+ kg).
1.71. (a) w1 โ^ ("11 โ "12) g-1- 2m2wo ยฑ m
(b) F = rn4:+ 1 17: (gโ w0).
1.72. w= 2g (2i โ sina)/(41 + 1). 1.73. (^) = 4mini2 -1-mo (m1โ ins) (^1) 4m1m2+ mo (^) + m2) 5. 1.74. Fir= 21mMI(M โ m) t2. 1.75. t =1/2/ (4 ยฑ Ti)/3g (2โ r() 1.4 s. 1.76. H = /(1-1 + 4) = 0.6 m.
= (^) ml (^7) 17m2 2 (g w0):
Fig. 4. Fig. 5.
1.77. WA = gl(1 cote a),^ w B = g/(tan a + cot a). 1.78. w= g V D(2+^ k+ 1.79. wmia = g (1 โ k)/(1 k). 1.80. wmax = g (1 k cot a)/(cot a โ^ k). 1.81. w = g sin a cos a/(sin2a^ ml/m2). 1.82. w โ
mg sin a M 2m (1โ cos a) ' 1.83. (a) l(F)1= 2 -1/ 2 mv2htR;^ (b) l(F)1=^ mac. 1.84. 2.1, 0.7 and 1.5 kN. 1.85. (a) w V1+ 3 cos20, T =3mg cos 0; (b) T= mg j/3; (c) cos 0=1/V "g, = 54.7ยฐ. 1.86. 53ยฐ. 1.87. 0 = arccos (2/3) ^ 48ยฐ, v = V 2gR/3. 1.88. a = 1/(x/nuo2โ 1). Is independent of the rotation di- rection. 1.89. r =R/2, (^) vmax = 1/2 kgR. 1.90. s = 1/2R V (kg/w.02โ1 =^ 60 m. 1.91. v C a V kg/a. 1.92. T = (cot 0 + co2RIg) mg/2n. 1.93. (a) Let us examine a small element of the thread in contact with the pulley (Fig. 5). Since the element is weightless, dT = dF j,. = k dFโ and^ dFโ = T da.^ Hence,^ dTIT = k^ da. Integrat-
ing this equation, we obtain k^ (ln no)/n; (b) w=^ g โ lovoi +TO.
1.94. F = (mvโ21R) cost a. 1.95. F = โmco2r, where r is the radius vector of the particle relative to the origin of coordinates;
F - m (1)2V x2 + y2. 1.96. (a) Ap = mgt; (b) I Ap I = โ2m(vog)Ig. 1.97. (a) p = at3/6; (b) s = =--a-r4/12m. 1.98. s = (cot โ sin cot) Fo/ma)2, see Fig. 6. 1.99. t = n/o); s = 2F^ 01m(03; vmax = Foinuo. 1.100. (a) v=voe-trlm,^ 00;^ (b)^ v=vo โsrlm,
' 1.101. t =
h (vo โ v) vov In (vo/u) ' 1.102. s = โ a^2 tan a, vmax = 171a- 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 โ to)31m, where to = kmgla is the moment of time at which the motion starts. At t < tothe distance is^ s =^ 0. 1.104. v' = vo/ kvVmg. 1.105. (a) v = (2F/mw)^ I sin (o)t/2)I; (b) As = 8Firnco2, (v) = 4F/nnuo. 1.106. v = vo/(1 + cos cp). Instruction. Here w,^ โwx, and therefore v = โvx^ const. From the initial condition it follows that const = vo. Besides, vx = v cos q. 1.107. w = [1 โ cos (11R)] Rgll.
1.108. (a) v = -1/2gfl /3; (b) cos % =
3 (1+ 2 )^ '^ where 1= = wo/g, 00 17ยฐ. 1.109. For n <1, including negative values. 1.110. When c02R > g, there are two steady equilibrium posi- tions: 01= 0 and 02= arccos^ (g1(02R).^ When OR^ <g,^ there is only one equilibrium position: 01= 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 ((os21v) sin cp = 7 cm, where 0) is the angular veloc- ity of the Earth's rotation. 1.112. F = m g2 w4r 2ยฑ (211(0)2 = 8 N. 1.113. Feor = 2m0)2r 1/1 (vohor)2----- 2.8 N.
Off (^) fit ad Fig. 6.
Stotal
1.146. (al (^) Fir =mg [sin a + (w211 g) ccs al= (^) 6 N. (b) co< <V g (kโ^ tan a)// (1^ k^ tan cz) = 2 rad/s. 1.147. (a) V = (mivi m2v2)/(m1+ m2); (b) T = ฮผ (v 1 โ โ v2)212, where p, = m1m2 (ml-h m2). 1.148. E E mV2/2. 1.149. E = -I- v22)/2, where p. = mi.m2/(m1 m2). 1.150. p = Po mgt, where pc, = mvi m2v2, m = (^) m2; re = vot gt2/2, where v0= (m1v1 m2v2)/(ml (^) m2)โข 1.151. ve = xV (^) xm2/(mi. ยฑ m2). 1.152. (a) /max = /0 Flx, (^) can = lo; (b) /max = (^) to + 2m1F/x (m1 + m2), /min = 4,โข 1.153. (a) Al > 3mg/x; (b) h = (1 + xAl/mg)2mg/8x = 8mg/x. 1.154. v1= โmv/(M โ m), v2 = Mv/(M โ , mM 1.155. vrear vo โ u; vform = vo -t- +no,^ U.
1.156. (1) viโ โ m+2m u;^ 2m (2) v2 โ (m+m 2M + 3m)(m)(m+ 2m) u,
V2k, = 1 -^1 -^ m/2 (M -1- m) >^ 1. 1.158. Ap = m y 2gh 1)/(71โ 1) = 0.2 kg -m/s. 1.159. (a) Iโ
m
jw+m I '; (b) Fโ
mM dv' M m dt โข 1.160. 1 = m1'12M. 1.161. -c= (p cos aโ M 17-2g1 sin a)/Mg sin a. 1.162. (a) v = (2M/m) Vir sin (0/2); (b) 1โ m/M. 1.163. h = Mv2/2g (M m). 1.164. (1) A = โRgh,^ where p, =^ mM/(m M);^ (2) Yes. 1.166. v = 1.0i + 2.0j โ 4.0k, v 4.6 m/s. 1.167. AT = --p (v1โ v2)212, where p, = m1m2/(m1 m2). 1.168. (a) 11 = 2m (m
-I-m2); (b)^ =^ 4m^1 m^2 /(m^1 +^ m2)2- 1.169. (a) m 1 /m2= 1/3; (b) m1/m2= 1 + 2 cos e = 2.0. 1.170. 11 = 1/2cos2 a = 0.25. 1.171. (^) max= v (1 + -I/2 (1- 1)) =1.0 km per second. 1.172. Will continue moving in the same direction, although this time with the velocity v' = (1โ V1 โ 2i) v/2. For 11< 1 the velocity v' Tiv/2 = 5 cm/s. 1.173. AT IT = (1 mlM)^ tang 0^ m/M^ โ 1 =^ โ^ 40%. 1.174. (^) (a) p = IAA / 14. v:; (b) T '/211(v - Fv:). Here p.= = mim2/(mi+ m2)- 1.175. Sin 0max 1.176. v' = โv (2 โ r12)/(6 โ r12). Respectively at smaller 11,
equal, or greater than V T. 1.178. Suppose that at a certain moment t the rocket has the mass m and the velocity v relative to the reference frame employed. Consider the inertial reference frame moving with the same velocity as the rocket has at a given moment. In this reference frame the momentum increment that the system "rocket-ejected portion of gas"
acquires during the time dt is equal to dp = m dv (^) dtโขu = F dt. What follows is evident. 1.179. v = โu In (mo/m). 1.180. m = moe-wt /u. 1.181. a = (u/vo) In (mo/m). 1.182. v = โ
F In m0 w (^) Rt โข 11 me 1.183. v = Ft/m0(1 (^) tp,t/m0), w
nzoโ = F/m0(1 (^) Rtimor- 1.184. v (^) 2gh ln (11h). 1.185. N=2b V alb. 1.186. M = i/2mgvot2cos a; M = (mv:/2g) sine a cos a = = 37 kg โข m2/s. 1.187. (a) Relative to all points of the straight line drawn at right angles to the wall through the point 0; (b) I AM I = 2 mv/ cos a. 1.188. Relative to the centre of the circle. I AM I = 2 V1 โ (g/&21)2 mgl/w. 1.189. I AM I = hmV. 1.190. M = maivgt2. 1.191. m = 2kr1/v22. 1.192. vo =1/-2g//cos O. 1.193. F = mcogr:Ir3. 1.194. M Z= Rmgt. 1.195. M = Rmgt sin a. Will not change. 1.196. M' = M โ [rot)]. In the case when p = 0, i.e. in the frame of the centre of inertia. 1.198. M = 1/3imvo. 1.199. Erna. mil/x102. The problem is easier to solve in the frame of the centre of inertia. 1.200. T = 23-iyM/v3= 225 days. 1.201. (a) 5.2 times; (b) 13 km/s, 2.2.10-4m/s2. 1.202. T = (r R)3/2yM. It is sufficient to consider the motion along the circle whose radius is equal to the major semi-axis of the given ellipse, i.e. (r R)12, since in accordance with Kepler's laws the period of revolution is the same. 1.203. Falling of the body on the Sun can be considered as the motion along a very elongated (in the limit, degenerated) ellipse whose major semi-axis is practically equal to the radius R of the Earth's orbit. Then from Kepler's laws, (2T/T)2 = (^) [(R12)1R13, where i is the falling time (the time needed to complete half a revo- lution along the elongated ellipse), T is the period of the Earth's revolution around the Sun. Hence, T = T/4112 = 65 days. 1.204. Will not change. 1.205. 1= f y M (T/231)2. 1.206. (a) U = โ ym1m2/r; (b) U = โy (mM11) In + 11a); F yrnM I a (a +1). 288
