Integrating these equations, with the initial conditions taken into account, we get x =--- (volco) (1 — cos cot), y = (vo/o)) sin cot. Hence (v0/(10. vo /(.0)2 + y2 ) This is the equation of a circle of radius vo/co with the centre at the point xo = vo/o), yo = 0. 4.43. Will increase V 1 + 2/5 (R//)2 times. It is taken into account here that the water (when in liquid phase) moves translationwise, and the system behaves as a mathematical pendulum. 4.44. co =
2x/
2/
(1+ — mg ) •
4.45. (a) T == 2n y 113g =1.1 s; (b) E = il2mg/a2 = 0.05 J. 4.46. (pm= cpo l/ 1 + mR2cp/2kcp:, E= 1/2ken. 4.47. (T) = 118mg10: V12702620. 4.48. 7' = 4n/a). 0.8 g m 2. 4.49. I = m12(co:— g11)1(0, — 04) 4.50. co = I/(Iico; + /2o)22 )/(/1 + /2).• 4.51. x= 112V, T min = 2n 1/11gVS. 4.52. T = n V 2h/g, 1„d = h/2. 4.53. coo = / 3(1102/2/. 4.54. coo =-1/ x/(m //R2). 4.55. coo =
2mg cos a MR +2mR (1+ sin a) '
4.56. T 2n1/ 3(R—r)12g. 4.57. T = ni/ 3m/2x. 4.58. co, = V x/R, where IA = m1m2/(m1± m2). 4.59. (a) = 1/ x/p, = 6 s-i ; (b) E= 112p.v2i = 5 mJ, a=vileo--= 2 cm. Here p, = m1m2/(m1 + m2). 4.60. T = 2n y rk, where I' =11121(11 + /2). 4.61. co2/co1 V1 2mo/mc 1.9, where mo and me are the masses of oxygen and carbon atoms. 4.62. co = 81/21,,po/mVo , where Y is the adiabatic exponent. 4.63. q= 4h1/ acomg (i2 — 1) = 2.0 p,C. 4.64. The induction of the field increased 112 = 25 times. 4.65. x = (vo/co) sin cot, where w = 1B/V mL. 4.66. x —(1—cos wt) g/w2, where w = /B/VmL. (arctan +nn) , where 4.67. (a) cto and now; (b) to (3 n=0, 1, 2, ... cp (0) = (32— (1)2)% (b) t r, = 4.68. (a) ip(0) = — 13cpo, 0,2_ R2 r -Frin), where n = 0, 1, 2, = (arctan
=w
o.)
2p(o