Irodov – Problems in General Physics

Integrating these equations, with the initial conditions taken into
account, we get x =--- (volco) (1 — cos cot), y = (vo/o)) sin cot. Hence
vo/ (.0)2 + y2^ (v0 / (1^0 ) 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 translation-
wise, and the system behaves as a mathematical pendulum.

4.44. co = (^) 2/ (1+ —mg ) •
2x/

4.45. (a) T == 2n y 113g =1.1 s; (b) E = il^2 mg/a^2 = 0.05 J.

4.46. (pm = cpo l/ 1 + mR 2 cp/2kcp:, E=^1 /^2 ken.

4.47. (T) = 118 mg10: (^) V12 70262 0.
4.48. 7' = 4n/a).
4.49. I = m1 2 (co:— g11)1(0, —^04 )^ 0.^8 g m^2.^
4.50. co = I/(Iico; + / 2 o) 22 )/(/^1 + /^2 ).•
4.51. x= 112V, Tmin= 2n 1/11gVS.
4.52. T = n V 2h/g, 1„d =

4.53. coo = / 3(110 2 /2/.

4.54. coo = -1/ x/(m //R 2 ).

2mg cos a

4.55. coo = (^) 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 = m^1 m^2 /(m^1 ± m^2 ).
4.59. (a) = 1/ x/p, = 6 s-i ; (b) E=^112 p.v^2 i = 5 mJ, a=vileo--=
2 cm. Here p, = m^1 m^2 /(m^1 + m^2 ).

4.60. T = 2n y rk, where I' =1^112 1(1^1 + /^2 ).

4.61. co 2 /co 1 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 (i^2 — 1) = 2.0 p,C.

4.64. The induction of the field increased 11^2 = 25 times.

4.65. x = (vo/co) sin cot, where w = 1B/V mL.

4.66. x —(1—cos wt) g/w^2 , where w = /B/VmL.

4.67. (a) cto and now; (b) t = wo (arctan (3 + nn) , where

n=0, 1, 2, ...

4.68. (a) ip (0) = — 13cpo ,^ cp (0) = (3^2 — (1)^2 )% (b)^ t r, =

0 ,2_ R2

=o.) (arctan (^) 2p(or -Frin), where n = 0, 1, 2,
h/2.