18.11: Kasi ya Angular

Angalia Uelewa Wako

11.1. a.\(\mu_{S} \geq \frac{\tan \theta}{1 + \left(\dfrac{mr^{2}}{I_{CM}}\right)}\); kuingiza angle na kubainisha kuwa kwa silinda mashimo I _CM = mr ², tuna\(\mu_{S} \geq \frac{\tan 60^{o}}{1 + \left(\dfrac{mr^{2}}{mr^{2}}\right)} = \frac{1}{2} \tan 60^{o}\) = 0.87; tunapewa thamani ya 0.6 kwa mgawo wa msuguano wa tuli, ambayo ni chini ya 0.87, hivyo hali haijastahili na silinda ya mashimo itaingizwa; b. the silinda imara inatii hali\(\mu_{S} \geq \frac{1}{3} \tan \theta = \frac{1}{3} \tan 60^{o}\) = 0.58. Thamani ya 0.6 kwa\(\mu_{S}\) ajili ya kukidhi hali hii, hivyo silinda imara haiwezi kuingizwa.

11.2. Kutoka kwenye takwimu, tunaona kwamba bidhaa ya msalaba ya vector ya radius na vector ya kasi inatoa vector iliyoongozwa nje ya ukurasa. Kuingiza radius na kasi katika kujieleza kwa kasi ya angular, tuna $$\ vec {l} =\ vec {r}\ mara\ vec {p} = (0.4\; m\;\ kofia {i})\ mara [1.67\ mara 10^ {-27}\; kg (4.0\ mara 10^ {6}\; m/s)\ kofia {j}] = 2.7\ mara 10^ {-21}\; kilo\ cdotp m^ {2} /s\;\ kofia {k}\]

11.3. I _nyanja =\(\frac{2}{5}\) mr ², I _silinda =\(\frac{1}{2}\) mr ²; Kuchukua uwiano wa momenta angular, tuna:\(\frac{L_{cylinder}}{L_{sphere}} = \frac{I_{cylinder} \omega_{0}}{I_{sphere} \omega_{0}} = \frac{\frac{1}{2} mr^{2}}{\frac{2}{5} mr^{2}} = \frac{5}{4}\). Hivyo, silinda ina kasi zaidi ya 25% ya angular. Hii ni kwa sababu silinda ina molekuli zaidi kusambazwa mbali na mhimili wa mzunguko.

11.4. Kutumia uhifadhi wa kasi ya angular, tuna mimi (4.0 rev/min) = 1.25I\(\omega_{f}\),\(\omega_{f}\) =\(\frac{1.0}{1.25}\) (4.0 rev/min) = 3.2 rev/min

11.5. Mvuto wa Mwezi ni 1/6 ule wa Dunia.Kwa kuchunguza Equation 11.12, tunaona kwamba mzunguko wa juu wa precession ni linearly sawia na kuongeza kasi ya mvuto. Wengi wengine wote, wingi, wakati wa inertia, na kiwango cha spin ni sawa na Mwezi. Hivyo, mzunguko wa precession juu ya Mwezi ni\(\omega_{P}\) (Mwezi) =\(\frac{1}{6} \omega_{P}\) (Dunia) =\(\frac{1}{6}\) (5.0 rad/s) = 0.83 rad/s.

Maswali ya dhana

1. Hapana, nguvu ya msuguano wa tuli ni sifuri.

3. Gurudumu ni zaidi ya kuingizwa kwenye mwelekeo mwinuko, kwani mgawo wa msuguano wa tuli lazima uongeze kwa angle ili kuendelea na mwendo usiozunguka bila kuacha.

5. Silinda hufikia urefu mkubwa. Kwa Equation 11.20, kuongeza kasi yake katika mwelekeo chini ya kutembea itakuwa chini.

7. Pointi zote kwenye mstari wa moja kwa moja zitatoa kasi ya angular, kwa sababu vector ilivuka kwenye vector sambamba ni sifuri.

9. Chembe lazima iongozwe kwenye mstari wa moja kwa moja unaopita kupitia asili iliyochaguliwa.

11. Bila propeller ndogo, mwili wa helikopta ungekuwa mzunguko kwa maana kinyume na propeller kubwa ili kuhifadhi kasi ya angular. Propeller ndogo hutoa msukumo kwa umbali R kutoka katikati ya wingi wa ndege ili kuzuia hili kutokea.

13. Kasi ya angular huongezeka kwa sababu wakati wa inertia unapungua.

15. Masi zaidi hujilimbikizia karibu na mhimili wa mzunguko, ambayo inapungua wakati wa inertia na kusababisha nyota kuongeza kasi yake ya angular.

17. Wakati unahitajika katika mwelekeo perpendicular kwa vector kasi angular ili kubadilisha mwelekeo wake. Majeshi haya kwenye gari la nafasi ni nje ya chombo ambacho gyroscope imewekwa na haitoi torques kwenye disk inayozunguka ya gyroscope.

Matatizo

19. v _CM = R\(\omega \Rightarrow \omega\) = 66.7 rad/s

21. \(\alpha\)= 3.3 rad/s ²

23. I _CM =\(\frac{2}{5}\) mr ², _CM = 3.5 m/s ²; x = 15.75 m

25. Chanya ni chini ya ndege ya kutembea; $$a_ {CM} =\ frac {mg\ dhambi\ theta} {m +\ kushoto (\ dfrac {I_ {CM}} {r^ {2}}\ haki)}\ Rightarrow I_ {CM} = r^ {2}\ Bigg [\ frac {mg\ dhambi 30} {a_ {CM}}}], $$$$x - x_ {0} = v_ {0} t -\ frac {1} {2} a_ {CM} t^ {2}\ Rightarrow a_ {CM} = 2.96\; m/s^ {2}, $$$I_ {CM} = 0.66\ mr ^ {2}\]

27. \(\alpha\)= 67.9 rad/s ², (_cm) x = 1.5 m/s ²

29. W = -1080.0 M

31. Nishati ya mitambo chini inalingana na nishati ya mitambo hapo juu; $$\ frac {1} {2} mv_ {0} ^ {2} +\ frac {1} {2}\ kushoto (\ dfrac {1} {2} mr^ {2}\ haki)\ kushoto (\ dfrac {0}} {r}\ haki) ^ {2} = mgh\ Rightarc Row h =\ frac {1} {g}\ kushoto (\ dfrac {1} {2} +\ dfrac {1} {4}\ haki) v_ {0} ^ {2}, $$h = 7.7 m, hivyo umbali juu ya kutembea ni 22 0.5 m.

33. Matumizi ya uhifadhi wa nishati $$\ kuanza {split}\ frac {1} {2} mv_ {0} ^ {2} +\ Frac {1} {2} I_ {Cyl}\ omega_ {0} ^ {2} & = mgh_ {Cyl},\\\ Frac {1} {2} mv_ {0} ^ {2} +\ Frac {1} {2} I_ {Sph}\ omega_ {0} ^ {2} & = mgh_ {Sph}\ ldotp\ mwisho {mgawanyiko} $$Kutoa milinganyo miwili, kuondoa nishati ya awali ya kutafsiri, tuna $$\ frac { 1} {2} I_ {Cyl}\ omega_ {0} ^ {2} -\ frac {1} {2} {2} I_ {Sph}\ omega_ {0} ^ {2} = mg (h_ {Cyl}} - h_ {Sph}), $$$$\ frac {1} {2} mr^ {2}\ kushoto (\ dfrac {vrac} _ {0} {r}\ haki) ^ {2}}\ frac {1} {2}\ kushoto (\ dfrac {2} {2} {3}\ haki) mr^ {2}\ kushoto (\ dfrac {v_ {0}} {r}\ haki) ^ {2} = mg (h_ {Cyl}} - h_ {Sph}), $$$$\ fra c {1} {2} v_ {0} ^ {2}}}\ frac {1} {2}\ kushoto (\ dfrac {2} {3}\ haki) v_ {0} ^ {2} = g (h_ {Cyl} - h_ {Sph}), $$$$h_ {Cyl} - h_ {Sph} =\ frac {1} {g}\ kushoto (\ dfrac {1} {2}} {2}}\ dfrac {1} {3}\ haki) v_ {0} ^ {2} =\ Frac {1} {9.8\; m/s^ {2}}\ kushoto (\ dfrac {1} {6}\ haki) (5.0\; m/s) ^ {2} = 0.43\; m\ ldOTP $Kwa hiyo, nyanja ya mashimo, na wakati mdogo wa inertia, inaendelea hadi urefu wa chini wa 1.0 ÷ 0. 43 = 0.57 m.

35. Ukubwa wa bidhaa msalaba wa radius kwa ndege na vector yake ya kasi huzalisha rp dhambi\(\theta\), ambayo inatoa r dhambi\(\theta\) kama urefu wa ndege h. mwelekeo wa kasi angular ni perpendicular radius na kasi wadudu, ambayo sisi kuchagua kiholela kama\(\hat{k}\), ambayo iko katika ndege ya ardhi: $$\ vec {L} =\ vec {r}\ nyakati\ vec {p} = hmv\;\ kofia {k} = (300.0\; m) (2.0\; kg) (20.0\; m/s)\;\ kofia {k} = 12,000.0\; kg\ cdotp m^ {2} /s\;\ kofia {k} = 12,000.0\; kg\ cdotp m^ {2} /s\;\ kofia {k} = 12,000.0\; kg\ cdotp m^ {2} /s\;\ kofia kofia {k}\]

37. a.\(\vec{l}\) = 45.0 kg • m ² /s\(\hat{k}\)

b.\(\vec{\tau}\) = 10.0 N • m\(\hat{k}\)

39. a.\(\vec{l}_{1}\) = -0.4 kg • m ² /s\(\hat{k}\),\(\vec{l}_{2} = \vec{l}_{4}\) = 0,\(\vec{l}_{3}\) = 1.35 kg • m ² /s\(\hat{k}\)

b.\(\vec{L}\) = 0.95 kg • m ² /s\(\hat{k}\)

41. a. L = 1.0 x 10 kilo ¹¹ • m ² /s

b Hapana, kasi ya angular inakaa sawa tangu bidhaa ya msalaba inahusisha tu umbali wa perpendicular kutoka ndege hadi chini bila kujali wapi iko njiani.

43. a.\(\vec{v} = −gt\; \hat{j}, \vec{r}_{\perp} = −d;\ \hat{i}, \vec{l} = mdgt\; \hat{k}\)

b.\(\vec{F} = −mg\; \hat{j}, \Sigma \vec{\tau} = dmg\; \hat{k}\)

c Ndiyo

45. a. mgh =\(\frac{1}{2}\) m (r\(\omega\)) ² +\(\frac{1}{2} \left(\dfrac{2}{5}\right) mr^{2} \omega^{2}\);\(\omega\) = 51.2 rad/s; L = 16.4 kg • m ² /s

b.\(\omega\) = 72.5 rad/s; L = 23.2 kg • m ² /s

47. a. mimi = 720.0 kg • m ²;\(\alpha\) = 4.20 rad/s ²;\(\omega\) (10 s) = 42.0 rad/s; L = 3.02 x 10 ⁴ kg • m ² /s;\(\omega\) (20 s) = 84.0 rad/s

b.\(\tau\) = 3.03 x 10 ³ N • m

49. a. L = 1.131 x 10 kilo ⁷ • m ² /s

b.\(\tau\) = 3.77 x 10 ⁴ N • m

51. \(\omega\)= 28.6 rad/s\(\Rightarrow\) L = 2.6 kg • m ² /s

53. $L_ {f} =\ frac {2} {5} M_ {S} (3.5\ mara 10^ {3}\; km) ^ {2}\ frac {2\ pi} {T_ {f}}, $$$ (7.0\ mara 10^ {5}\; km) ^ {2}\ frac {2\ pi} {2\ pi}} = (3.5\ mara 10^ {3}\; km) ^ {2}\ frac {2\ pi} {T_ {f}} $$$$T_ {f}\ Rightarrow = 28\; siku\ frac {(3.5\ mara 10^ {3}\; km) ^ {2}\; km) ^ {2}\ mara 10^ {-4}\; siku = 60.5\; s\]

55. f _f = 2.1 rev/s\(\Rightarrow\) f ₀ = 0.5 rev/s

57. _{r P mv P = r A mv\(\Rightarrow\) V P = 18.3 km/s}

59. a. mimi _disk = 5.0 x ^{10 -4} kg • m ², mimi _mdudu = 2.0 x 10 ^-4 kg • m ², (I _disk + mimi _mdudu)\(\omega_{1}\) = Mimi _disk\(\omega_{2}\),\(\omega_{2}\) = 14.0 rad/s

b.\(\Delta\) K = 0.014 J

c.\(\omega_{3}\) = 10.0 rad/s nyuma ya thamani ya awali

d.\(\frac{1}{2}\) (Mimi _disk + mimi _mdudu)\(\omega_{3}^{2}\) = 0.035 J nyuma thamani ya awali

e Kazi ya mdudu kutambaa kwenye diski

61. L _i = 400.0 kg • m ² s, L _f = 500.0 kg • m ²\(\omega\),\(\omega\) = 0.80 rad/s

63. I ₀ = 340.48 kg • m ², I _f = 268.8 kg • m ²,\(\omega_{f}\) = 25.33 rpm

65. a. L = 280 kg • m ² /s, I _f = 89.6 kg • m ²,\(\omega_{f}\) = 3.125 rad/s

b K _i = 437.5 J, K _f = 437.5 J

67. Muda wa inertia katika rekodi ya rekodi: I ₀ = 0.5 kg • m ², I _f = 1.1 kg • m ²,\(\omega_{f} = \frac{I_{0}}{I_{f}} \omega_{0} \Rightarrow \) f _f = 155.5 rev/min

69. Kiwango chake cha spin katika hewa ni: _{f f} = 2.0 rev/s; Anaweza kufanya flips nne katika hewa.

71. Muda wa inertia na watoto wote ndani: I ₀ = 2.4 x 10 kilo ⁵ • m ²; Mimi _f = 1.5 x 10 kilo ⁵ • m ²; f _f = 0.3 rev/s

73. I ₀ = 1.00 x 10 kilo ¹⁰ • m ², I _f = 9.94 x 10 ⁹ kg • m ², f _f = 3.32 rev/min

75. I = 2.5 x 10 ^-3 kg • m ²,\(\omega_{P}\) = 0.78 rad/s

77. a. L _Dunia = 7.06 x 10 kilo ³³ • m ² /s,\(\Delta\) L = 5.63 x 10 ³³ kg • m ² /s

b.\(\tau\) = 1.7 x 10 ²² N • m

c Vikosi viwili katika ikweta vingekuwa na ukubwa sawa lakini mwelekeo tofauti, moja katika mwelekeo wa kaskazini na nyingine upande wa kusini upande wa pili wa Dunia. Pembe kati ya vikosi na silaha za lever hadi katikati ya Dunia ni 90°, hivyo moment iliyotolewa ingekuwa na ukubwa\(\tau\) = FR _E dhambi 90° = FR _E. Wote watatoa wakati katika mwelekeo huo:\(\tau\) = 2FR _E\(\Rightarrow\) F = 1.3 x 10 ¹⁵ N

Matatizo ya ziada

79. _cm = -\(\frac{3}{10}\) g, v ² = v ₀ ² + 2a _cm x\(\Rightarrow\) v ² = (7.0 m/s) ² - 2\(\left(\dfrac{3}{10}g\right)\) x, v ² = 0\(\Rightarrow\) x = 8.34 m

b. t =\(\frac{v − v_{0}}{a_{CM}}\), v = v ₀ + _cm t\(\Rightarrow\) t = 2.38 s; Eneo la mashimo lina muda mkubwa wa inertia, na kwa hiyo ni vigumu kuleta mapumziko kuliko marumaru, au nyanja imara. Umbali uliosafiri ni mkubwa na muda uliopita ni mrefu.

81. a. W = -500.0 J

b K + U _grav = mara kwa mara, 500 J + 0 = 0 + (6.0 kg) (9.8 m/s ²) h, h = 8.5 m, d = 17.0 m; Wakati wa inertia ni mdogo kwa nyanja ya mashimo, kwa hiyo kazi ndogo inahitajika kuacha. Vivyo hivyo ni Rolls up elekea umbali mfupi kuliko hoop.

83. a.\(\tau\) = 34.0 N • m

b. l = mr ²\(\omega \Rightarrow \omega\) = 3.6 rad/s

85. a. d _M = 3.85 x 10 ⁸ m wastani wa umbali wa Mwezi; kipindi cha orbital 27.32d = 2.36 x 10 ⁶ s; kasi ya Mwezi\(\frac{2 \pi 3.85 \times 10^{8}\; m}{2.36 \times 10^{6}\; s}\) = 1.0 x 10 ³ m/s; wingi wa Mwezi 7.35 x 10 ²² kg, L = 2.90 x 10 ³⁴ kgm ² /s

b Radi ya Mwezi 1.74 x 10 ⁶ m; kipindi cha orbital ni sawa na (a):\(\omega\) = 2.66 x 10 ^-6 rad/s, L = 2.37 x 1029 kg • m ² /s; Kasi ya angular ya orbital ni 1.22 x 10 mara ⁵ kubwa kuliko kasi ya mzunguko wa angular kwa Mwezi.

87. I = 0.135 kg • m ²,\(\alpha\) = 4.19 rad/s ²\(\omega = \omega_{0} + \alpha t\),\(\omega\) (5 s) = 21.0 rad/s, L = 2.84 kg • m ² /s,\(\omega\) (10 s) = 41.9 rad/s, L = 5.66 kg • m/s ²

89. Katika uhifadhi wa equation ya kasi ya angular, kiwango cha mzunguko kinaonekana pande zote mbili kwa hiyo tunaweka alama (rev/min) kama kasi ya angular inaweza kuongezeka kwa mara kwa mara kupata (rev/min): L _i = -0.04 kg • m ² (300.0 rev/min), L _f = (0.08 kg • m ²) f _f\(\Rightarrow\) f _f = -150.0 rev/min saa moja kwa moja

91. I ₀\(\omega_{0}\) = I _f\(\omega_{f}\), I ₀ = 6120.0 kg • m ², I _f = 1180.0 kg • m ²,\(\omega_{f}\) = 31.1 rev/min

93. L _i = 1.00 x 10 kilo ⁷ • m ² /s, I _f = 2.025 x 10 kilo ⁵ • m ²,\(\omega_{f}\) = 7.86 rev/s

Changamoto Matatizo

95. Kudhani roll kuchochea kasi mbele kwa heshima na ardhi na kuongeza kasi a. Kisha huharakisha nyuma jamaa na lori na kuongeza kasi (a - a).

Pia, R\(\alpha\) = a - a, I =\(\frac{1}{2}\) mR ²,\(\Sigma\) F _x = f _s = ma,\(\Sigma \tau\) = f _s R = I\(\alpha\) = I\(\frac{a − a′}{R}\), f _s =\(\frac{I}{R^{2}}\) (a - a) =\(\frac{1}{2}\) m (a - a)

Kutatua kwa a: f _s =\(\frac{1}{2}\) m (a -a), a =\(\frac{a}{3}\), x - x 0 = v ₀ t +\(\frac{1}{2}\) saa ², d =\(\frac{1}{3}\) saa ², t =, kwa hiyo\(\sqrt{\frac{3d}{a}}\), s = 1.5d

97. a. mvutano katika kamba hutoa nguvu centripetal kama kwamba T dhambi\(\theta\) = mr _\(\perp\)\(\omega^{2}\). Sehemu ya mvutano ambayo ni wima inapinga nguvu ya mvuto kama vile T cos\(\theta\) = mg. Hii inatoa T = 5.7 N. kutatua kwa r _\(\perp\)= 0.16 m Hii inatoa urefu wa kamba kama r = 0.32 m Katika\(\omega\) = 10.0 rad/s, kuna angle mpya, mvutano, na radius perpendicular kwa fimbo. Kugawanya equations mbili zinazohusisha mvutano ili kuondokana nayo, tuna\(\frac{\sin \theta}{\cos \theta} = \frac{(0.32\; m\; \sin \theta) \omega^{2}}{g} \Rightarrow \frac{1}{\cos \theta} = \frac{(0.32\; m) \omega^{2}}{g}\); cos\(\theta\) = 0.31\(\Rightarrow \theta\) = 72.2°

b. l _awali = 0.08 kg • m ² /s, l _mwisho = 0.46 kg• m ² /s

c Hapana, cosine ya angle ni inversely sawia na mraba wa kasi ya angular, kwa hiyo ili\(\theta\) → 90°,\(\omega\) → Δ. Fimbo ingekuwa na spin kwa haraka sana.

Wachangiaji na Majina

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