Irodov – Problems in General Physics

(Joyce) #1

1.16. tm = vilvf ' ± v212 (^) '
1.17. C D = I ji — 1.
1.18. See Fig. lb.
1.19.' (a) (v) aR/T = 50 cm/s; (b) 1(v)1 = 2R/r = 32 cm/s;
(c) j(w)1 = 2nR/T 2 = 10 cm/s 2.
1.20. (a) v = a(1 — 2at), w = —2aa = const; (b) At = 1/a,
s = a/2a.
1.21. (a) x = vot (1 —t/2t), x =0.24, 0 and — 4.0 m;
(1 — t/a) vot for t<__T, -
24
[1 4- (1 —^2 ] vot/2 for
4 and 34 cm respectively.
(b) 1.1, 9 and 11 s; (0) s =^ 0-0
1.22. (a) v = a 2 t/2, w = a 2 /2; (b) (v).= a V-s-/2.
1.23. (a) s = ( 2 / 3 a) vo/ 2 ; (b) t = 2V 1 ' 0 1 a.
1.24. (a) y = —x 2 b1a 2 ; (b) v = ai — 2btj, w = --2bj, v
1 1 a2 46 ,2 0 = 2b; (c) tan a = al2bt; (d) (v) = ai —btj, 1(0 =
= Va 2 + b 2 t 2.
1.25. (a) y = x — x (^2) a/a; (b) v = a -V 1 + (1 — 2at) 2 , w = 2aa
= const; (c) to = 1/a.
1.26. (a) s= aorr; (b)
1.27. vo = (1+ a 2 ) w/2b.
1.28. (a) r = v
°
t gt 2 /2; (b) (v ) = gcos a t
= vo gt/2, (v) = vo — g (vog)/g (^2) • ysin
1.29. (a) i = 2 (vo/g) sin a;
(b) h= (vV2g) sin 2 a, / = (v 2 o /g) sin 2a,
= 76'; r/2^ z t
(c) y = x tan a — (g/24 cost a) x2;
-11 sin a
(d) R-, = g cos a, R 2 = (vV g) cost a.
1.30. See Fig. 2.
1.31. / = 8h sin a.
1.32. 0.41 or 0.71 min later, depending on the initial angle.
1.33. At = 2vo^ sin (0^1 —^0 2)
g cos 0 1 + cos =11 s.
02
1.34. (a) x = (a/2V 0 ) y 2 ; (b) iv = avo, w-r = a 2 Y (^) (aY/vo) 2 '
wn = avo/V I + (ay/vo) 2 -
1.35. (a) (^) y = (b12a) x 2 ; (b) (^) R = v 2 /w 7 , = v 2 141 w 2
= (alb) [1 + (xbla)9 3/2.
1.36. v = y 2ax.
1.37. w= a V 1+ (43trt) 2 = 0.8 m/s 2.
1.38. (a) v = vo/(1 vot/R) = v fie -81R; (b) w= v 2 41Re 2 s/n=--
=- y2v2/R.
1.39. tan a = 2s1 R.
1.40. (a) wo = a 2 6) 2 I R = 2.6 m/s 2 , wa = aa) 2 = 3.2 m/s (^2) ; (b) Wm in
ao) 2 V 1 — (R/2a) 2 = 2.5 m/s 2 , lm = ± a V 1 — R 2 /2a 2 = ± 0.37 m.
lmrn—
1
1
v2-12vi
l v? +
Fig. 2.
282

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