18.10: Utangulizi wa mzunguko wa Axis
- Page ID
- 176682
Angalia Uelewa Wako
10.1. a. 40.0 rev/s = 2\(pi\) (40.0) rad/s,\(\bar{\alpha} = \frac{\Delta \omega}{\Delta t} = \frac{2 \pi (40.0) − 0\; rad/s}{20.0\; s}\) = 2\(\pi\) (2.0) = 4.0\(\pi\) rad/s 2
b Kwa kuwa kasi ya angular inaongezeka kwa mstari, kuna lazima iwe na kasi ya mara kwa mara wakati ulioonyeshwa. Kwa hiyo, kasi ya angular instantaneous wakati wowote ni suluhisho la 4.0\(\pi\) rad/s 2.
10.2. a Kutumia Equation 10.25, tuna 7000 rpm\(\frac{7000.0(2 \pi\; rad)}{60.0\; s}\) = = 733.0 rad/s, $$\ alpha =\ frac {\ omega -\ omega_ {0}} {t} =\ frac {733.0\; rad/s} {10.0\; s} = 73.3\; rad/s^ {2}\]
b Kutumia Equation 10.29, tuna $$\ omega^ {2} =\ omega_ {0} ^ {2} + 2\ alpha\ Delta\ theta\ Rightarrow\ Delta\ theta =\ frac {\ omega^ {2} -\ omega_ {0} ^ {2}} {2\ alpha} =\ frac {0 - (733.0\; rad/s) ^ {2}} {2 (73.3\; rad/s^ {2})} = 3665.2\; rad\]
10.3. Kuongeza kasi ya angular ni\(\alpha = \frac{(5.0 − 0)\; rad/s}{20.0\; s}\) = 0.25 rad/s 2. Kwa hiyo, angle ya jumla ambayo mvulana hupita ni $$\ Delta\ theta =\ frac {\ omega^ {2} -\ omega_ {0} ^ {2}} {2\ alpha} =\ frac {(5.0) ^ {2} - 0} {2 (0.25)} = 50\; Rad$$hivyo, tunahesabu $$s = r\ theta = (5.0\; m) (50.0\; rad) = 250.0\; m\]
10.4. Nishati ya awali ya mzunguko wa kinetic ya propeller ni $$K_ {0} =\ frac {1} {2} I\ omega^ {2} =\ frac {1} {2} (800.0\; kg\ cdotp m^ {2}) (4.0\ mara 2\ pi\; rad/s) ^ {2 = 2.53\ mara 10^ {5}\; J\ LdOTP $katika 5.0 s nishati mpya ya mzunguko wa kinetic ya propeller ni $$K_ {f} = 2.03\ mara 10^ {5}\; J\ ldotp $na kasi mpya ya angular ni $$\ omega =\ sqrt {\ frac {2 (2.03\ mara 10^ {5}\; J)} {800.0\; kg\ cdotp m^ {2}}} = 22.53\; rad/s $ambayo ni 3.58 rev/s.
10.5. Mimi sambamba-mhimili = Mimi katikati ya molekuli + md 2 = mR 2 + mR 2 = 2mR 2
10.6. Pembe kati ya mkono wa lever na vector nguvu ni 80°; kwa hiyo, r \(\perp\)= 100 m (dhambi 80°) = 98.5 m Bidhaa ya msalaba\(\vec{\tau} = \vec{r} \times \vec{F}\) hutoa wakati mbaya au wa saa. Wakati huo ni kisha\(\tau = - r_{\perp} F\) = -98.5 m (5.0 x 10 5 N) = -4.9 x 10 7 N • m.
10.7. a. kuongeza kasi ya angular ni\(\alpha = \frac{20.0(2 \pi)\; rad/s − 0}{10.0 \; s}\) = 12.56 rad/s 2. Kutatua kwa wakati, tuna\(\sum_{i} \tau_{i} = I \alpha\) = (30.0 kg • m 2) (12.56 rad/s 2) = 376.80 N • m
b Kuongeza kasi ya angular ni\(\alpha = \frac{0 − 20.0(2 \pi)\; rad/s}{20.0\; s}\) = -6.28 rad/s 2. Kutatua kwa wakati, tuna\(\sum_{i} \tau_{i} = I \alpha\) = (30.0 kg • m 2) (-6.28 rad/s 2) = -188.50 N • m
10.8. 3 MW
Maswali ya dhana
1. Mkono wa pili huzunguka saa moja kwa moja, hivyo kwa utawala wa mkono wa kulia, vector ya kasi ya angular iko ndani ya ukuta.
3. Wana kasi sawa ya angular. Pointi zaidi nje ya popo na kasi kubwa tangential.
5. Mstari wa moja kwa moja, mstari katika kutofautiana wakati
7. Mara kwa mara
9. Vector centripetal kasi ni perpendicular kwa vector kasi.
11. a. wote; b. nonzero centripetal kuongeza kasi; c. wote
13. Sehemu ya mashimo, kwani wingi unasambazwa zaidi mbali na mzunguko wa mzunguko.
15. a. itapungua. b. silaha inaweza kuwa takriban na fimbo na discus na diski. Torso iko karibu na mhimili wa mzunguko hivyo haichangia sana wakati wa inertia.
17. Kwa sababu wakati wa inertia hutofautiana kama mraba wa umbali na mhimili wa mzunguko. Masi ya fimbo iko katika umbali mkubwa kuliko L/2 ingekuwa kutoa mchango mkubwa wa kufanya wakati wake wa hali kubwa kuliko molekuli uhakika katika\(\frac{L}{2}\).
19. Ukubwa wa nguvu, urefu wa mkono wa lever, na angle ya mkono wa lever na vector nguvu
21. Wakati wa inertia ya magurudumu umepunguzwa, hivyo wakati mdogo unahitajika ili kuharakisha.
23. Ndiyo
25. |\(\vec{r}\) | inaweza kuwa sawa na mkono wa lever lakini sio chini ya mkono wa lever
27. Ikiwa majeshi ni pamoja na mhimili wa mzunguko, au ikiwa wana mkono sawa wa lever na hutumiwa kwa hatua kwenye fimbo.
Matatizo
29. \(\omega = \frac{2 \pi\; rad}{45.0\; s}\)= 0.14 rad/s
31. a.\(\theta = \frac{s}{r} = \frac{3.0\; m}{1.5\; m}\) = 2.0 rad
b.\(\omega = \frac{2.0\; rad}{1.0\; s}\) = 2.0 rad/s
c.\(\frac{v^{2}}{r} = \frac{(3.0\; m/s)^{2}}{1.5\; m}\) = 6.0 m/s 2.
33. Propeller inachukua tu\(\Delta\) t\(\frac{\Delta \omega}{\alpha} = \frac{0\; rad/s − 10.0(2 \pi)\; rad/s}{−2.0\; rad/s^{2}}\) = 31.4 s kuja kupumzika, wakati propeller ni saa 0 rad/s, ingeanza kupokezana katika mwelekeo kinyume. Hii haiwezekani kutokana na ukubwa wa vikosi vinavyohusika katika kupata propeller kuacha na kuanza kupokezana katika mwelekeo kinyume.
35. a.\(\omega\) = 25.0 (2.0 s) = 50.0 rad/s
b.\(\alpha = \frac{d \omega}{dt}\) = 25.0 rad/s 2
37. a.\(\omega\) = 54.8 rad/s
b. t = 11.0 s
39. a. 0.87 rad/s 2
b.\(\theta\) = 66,264 rad
41. a.\(\omega\) = 42.0 rad/s
b.\(\theta\) = 220 rad
c. v t = 42 m/s, t = 4.0 m/s 2
43. a.\(\omega\) = 7.0 rad/s
b.\(\theta\) = 22.5 rad
c. t = 0.1 m/s
45. \(\alpha\)= 28.6 rad/s 2.
47. r = 0.78 m
49. a.\(\alpha\) = -0.314 rad/s 2
b. c = 197.4 m/s 2
c. a\(\sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{197.4^{2} + (−6.28)^{2}}\) = 197.5 m/s 2,\(\theta\) = tan -1\(\frac{−6.28}{197.4}\) = -1.8° katika mwelekeo wa saa kutoka vector centripetal kuongeza kasi
51. ma = 40.0 kg (5.1 m/s 2) = 204.0 N
Nguvu ya msuguano wa juu ni\(\mu_{S}\) N = 0.6 (40.0 kg) (9.8 m/s 2) = 235.2 N hivyo mtoto haanguka bado.
53. $$\ kuanza {mgawanyiko} v_ {t} & = r\ omega = 1.0 (2.0t)\; m/s\\ a_ {c} & =\ frac {v_ {t} ^ {2}} {r} =\ frac {(2.0t) ^ {2}} {2}\ t {} (t) & = r\ alpha (t) = r\ frac {d\ omega} {dt} = 1.0\; m (2.0) = 2.0\; m/s^ {2}\ ldotp\ mwisho {mgawanyiko} $$Kupanga njama zote mbili inatoa
Kuongeza kasi ya kasi ni mara kwa mara, wakati kasi ya centripetal ni tegemezi ya wakati, na huongezeka kwa wakati wa maadili makubwa zaidi kuliko kasi ya tangential baada ya t = 1s. Kwa nyakati chini ya 0.7 s na inakaribia sifuri kasi ya centripetal ni kidogo sana kuliko kasi ya tangential.
55. a. K = 2.56 x 10 29 J
b. K = 2.68 x 10 33 J
57. K = 434.0 J
59. a. v f = 86.5 m/s
b. kiwango cha mzunguko wa propeller anakaa sawa katika 20 rev/s.
61. K = 3.95 x 10 42 J
63. a. mimi = 0.315 kg • m 2
b. K = 621.8 J
65. I =\(\frac{7}{36}\) mL 2
67. v = 7.14 m/s.
69. \(\theta\)= 10.2°
71. F = 30 N
73. a. 0.85 m (55.0 N) = 46.75 N • m
b Haijalishi kwa urefu gani unachochea.
75. m 2\(\frac{4.9\; N \cdotp m}{9.8(0.3\; m)}\) = 1.67 kg
77. \(\tau_{net}\)= -9.0 N • m + 3.46 N • m + 0 - 3.28 N • m = -8.82 N • m
79. \(\tau\)= 5.66 N• m
81. \(\sum \tau\)= 57.82 N • m
83. \(\vec{r} \times \vec{F}\)= 4.0\(\hat{i}\) + 2.0\(\hat{j}\) - 16.0\(\hat{k}\) N • m
85. a.\(\tau\) = (0.280 m) (180.0 N) = 50.4 N • m
b.\(\alpha\) = 17.14 rad/s 2
c.\(\alpha\) = 17.04 rad/s 2
87. \(\tau\)= 8.0 N • m
89. \(\tau\)= -43.6 N • m
91. a.\(\alpha\) = 1.4 x 10 -10 rad/s 2
b.\(\tau\) = 1.36 x 10 28 N • m
c F = 2.1 x 10 21 N
93. a = 3.6 m/s 2
95. a. a = r\(\alpha\) = 14.7 m/s 2
b. =\(\frac{L}{2} \alpha\) =\(\frac{3}{4}\) g
97. \(\tau = \frac{P}{\omega} = \frac{2.0 \times 10^{6}\; W}{2.1\; rad/s}\)= 9.5 x 10 5 N • m
99. a. K = 888.50 J
b.\(\Delta \theta\) = 294.6 rev
101. a. mimi = 114.6 kg • m 2
b. P = 104,700 W
103. v = L\(\omega\) =\(\sqrt{3Lg}\)
105. a. a = 5.0 m/s 2
b. W = 1.25 N • m
Matatizo ya ziada
107. \(\Delta\)t = 10.0 s
109. a. 0.06 rad/s 2
b.\(\theta\) = 105.0 rad
111. s = 405.26 m
113. a. mimi = 0.363 kg • m 2
b Mimi = 2.34 kg • m 2
115. \(\omega =\sqrt{\frac{5.36\; J}{4.4\; kg \cdotp m^{2}}}\)= 1.10 rad/s
117. F = 23.3 N
119. \(\alpha = \frac{190.0\; N \cdotp m}{2.94\; kg \cdotp m^{2}}\)= 64.4 rad/s 2
Changamoto Matatizo
121. a.\(\omega\) = 2.0t - 1.5t 2
b.\(\theta\) = t 2 - 0.5t 3
c.\(\theta\) = -400.0 rad
d. vector iko saa -0.66 (360°) = -237.6°
123. I =\(\frac{2}{5}\) mR 2
125. a.\(\omega\) = 8.2 rad/s
b.\(\omega\) = 8.0 rad/s