Isikhathi sokungahlaliseki kwezinhlayiya – izinkinga nezixazululo

Isikhathi sokungahlaliseki kwezinhlayiya - izinkinga nezixazululo

1. Amabhola amabili axhunywe yinduku, njengoba kuboniswe esithombeni esingezansi. Ungawanaki amanduku MassIsisindo sebhola u-P singamagremu angu-600 kanti isisindo sebhola u-Q singamagremu angu-400. Kuyini umzuzu we-inertia kohlelo mayelana ne-AB?

Kwaziwa:

I-axis yokujikeleza yi-AB. Isikhathi sokungahlaliseki kwezinhlayiya - izinkinga nezixazululo 1

m_p = amagremu angu-600 = 0.6 kg, m_q = amagremu angu-400 = 0.4 kg

r_p = 20 cm = 0.2 m, r_q = 50 cm = 0.5 m

Kufunwa: Isikhathi sokungahlali kahle kwesistimu

Isixazululo:

Mina = m_p r_p² +m_q r_q²

I = (0.6 kg)(0.2 m)² + (0.4 kg)(0.5 m)²

I = (0.6 kg)(0.04 m²) + (0.4 kg)(0.25 m²)

I = 0.024 kg m² + 0.1 kg m²

I = 0.124 kg m²

2. Induku ye-AB enesisindo esingu-2-kg ijikelezile cishe endaweni A, umzuzu wokungangeni kwenduku ungu-8 kg m²Uma kujikelezwe iphuzu O (AO = OB), yisiphi isikhathi sokungahlali kahle kwenduku.

Kwaziwa: Isikhathi sokungahlaliseki kwezinhlayiya - izinkinga nezixazululo 2

Isisindo senduku AB (m) = 2 kg

Uma kujikelezwe cishe iphuzu A ukuze irediyasi yokujikeleza (r) = ubude buka-AB = r khona-ke umzuzu we-inertia (I) = 8 kg m²

Okufunayo: Uma kujikelezwe cishe iphuzu O ukuze irediyasi yokujikeleza (r) = ubude be-AO = ubude be-OB = 1/2 r khona-ke yisiphi isikhathi sokungapheleli kwenduku.

Isixazululo:

Mina = uMnu.²

8 kg m² = (2 kg) r²

8 m² = (2) r²

r²= 8m² / 2

r²= 4m²

r = amamitha amabili

Uma kujikelezwe cishe iphuzu O ngakho ½ r = 1 imitha, khona-ke umzuzu we-inertia:

Mina = uMnu.²= (2 kg)(1 m)² = (2 kg)(1 m²) = 2 kg m²

3. Amabhola amabili axhunywe yinduku njengoba kuboniswe esithombeni esingezansi. Unganaki isisindo senduku. Iyini isikhathi sokungahlali kahle kwesistimu.

Kwaziwa:

Isisindo sebhola A (m_A) = amagremu angu-200 = 0.2 kg Isikhathi sokungahlaliseki kwezinhlayiya - izinkinga nezixazululo 3

Isisindo sebhola B (m_B) = amagremu angu-400 = 0.4 kg

Ibanga phakathi kwebhola A kanye ne-axis yokujikeleza (r)_A= = 0

Bhekafuthi Ukuqina komsindo - izinkinga nezixazululo

Ibanga phakathi kwebhola B kanye ne-axis yokujikeleza (r)_B) = 25 cm = 0.25 m

Kufunwa: Isikhathi sokungahlali kahle kwesistimu

Isixazululo:

Isikhathi sokungakhathali kwebhola A:

I_A = (m_A)(r_A²) = (0.2)(0)² = 0

Isikhathi sokungakhathali kwebhola B:

I_B = (m_B)(r_B²) = (0.4)(0.25)² = (0.4)(0.0625) = 0.025 kg m²

Isikhathi sokungabi namandla kwesistimu:

Mina = Mina_A+ Mina_B = 0 + 0.025 = 0.025 kg m² = 25 x10^-3 kg m²

4. Izinhlayiya ezine ezinesisindo esihlukene, eziboniswe esithombeni esingezansi. Nquma isikhathi sokungabi namandla kwesistimu mayelana nomugqa ovundlile P.

Isixazululo

I-axis yokujikeleza umugqa ovundlile u-P.

Kwaziwa: Isikhathi sokungahlaliseki kwezinhlayiya - izinkinga nezixazululo 4

Isisindo sezinhlayiya u-A (m_A) = m

Isisindo sezinhlayiya B (m_B) = 2m

Isisindo sezinhlayiya C (m_C) = 3m

Ukudlula kwenhlayiya D (m_D) = 4m

Ibanga phakathi kwe-particle A kanye ne-axis yokujikeleza (r)_A) = b

Ibanga phakathi kwe-particle B kanye ne-axis yokujikeleza (r)_B) = b

Ibanga phakathi kwe-particle C kanye ne-axis yokujikeleza (r)_C) = 2b

Ibanga phakathi kwe-particle D kanye ne-axis yokujikeleza (r)_D) = 2b

Kufunwa: Isikhathi sokungabi namandla kwesistimu mayelana nomugqa ovundlile P

Isixazululo:

Mina = m_A r_A² +m_B r_B²+m_C r_C²+m_Dr_D²

I = (m)(b)²+ (2m)(b)² + (3m)(2b)²+ (4m)(2b)²

I = mb²+ 2 mb² + (3m)(4b²) + (4m)(4b²)

I = mb²+ 2 mb² + 12 mb²+ 16 mb²

I = 31 mb²

5. Izinhlayiya ezine ezixhunywe yinduku. Unganaki isisindo senduku. Thola isikhathi sokungapheleli mayelana ne-axis yokujikeleza ngokusebenzisa inhlayiya m₁ kanye no-m₂, njengoba kuboniswe esithombeni esingezansi.

Kuyaziwa

Isisindo sezinhlayiya 1 (m₁) = 1/4 kg Isikhathi sokungahlaliseki kwezinhlayiya - izinkinga nezixazululo 5

Isisindo sezinhlayiya 2 (m₂) = 1/2 kg

Isisindo sezinhlayiya 3 (m₃) = 1/4 kg

Isisindo sezinhlayiya 4 (m₄) = 1/4 kg

Ibanga phakathi kwe-particle 1 kanye ne-axis yokujikeleza (r)₁= = 0

Ibanga phakathi kwe-particle 2 kanye ne-axis yokujikeleza (r)₂= = 0

Ibanga phakathi kwe-particle 3 kanye ne-axis yokujikeleza (r)₃) = 10 cm = 10/100 m = 1/10 m

Ibanga phakathi kwe-particle 4 kanye ne-axis yokujikeleza (r)₄) = 10 cm = 10/100 m = 1/10 m

Bhekafuthi Umsebenzi owenziwe ngenkani - izinkinga nezixazululo

Kufunwa: Isikhathi sokungakhathali

Isixazululo:

Mina = m₁ r₁² +m₂ r₂²+m₃ r₃²+m₄r₄²

I = (1/4)(0)²+ (1/2)(0)² + (1/4)(1/10)²+ (1/4)(1/10)²

I = 0 + 0 + (1/4)(1/100) + (1/4)(1/100)

I = 1/400 + 1/400

I = 2/400

I = 1/200 kg.m²

Iyini isikhathi sokungahlali kahle kwenhlayiya?
- Impendulo: Ngenhlayiya eyodwa yesisindo $m$ kude $r$ kusukela ku-axis of rotation, umzuzu wayo we-inertia $I$ unikezwa ngu $I = m \times r^{2}$ .
Kungani isikhathi sokungahlali isikhathi eside sivame ukubizwa ngokuthi “i-analogue yokujikeleza” yesisindo?
- ImpenduloNjengoba nje isisindo siyisilinganiso sokumelana kwento nezinguquko ekuhambeni kwayo kokuhumusha (ngenxa yomthetho wesibili kaNewton), umzuzu we-inertia uyisilinganiso sokumelana kwento nezinguquko ekuhambeni kwayo kokujikeleza.
Ukushintsha ibanga lenhlayiya kusuka ku-axis yayo yokujikeleza kuthinta kanjani umzuzu wayo we-inertia?
- Impendulo: Isikhathi sokungangeni silingana nesikwele sebanga ukusuka ku-axis yokujikeleza. Uma uphinda kabili ibanga, isikhathi sokungangeni sizokhula ngesilinganiso sesine.
Kungani isikwele sebanga (r) siyisikwele sebanga (r)²) esetshenziswe kufomula yesikhathi sokungahlali isikhathi eside esikhundleni sebanga nje?
- Impendulo: Isikwele sebanga sisetshenziswa ngenxa yendlela amandla e-kinetic ekujikelezeni asebenza ngayo. Ekuhambeni kokujikeleza, inhlayiya ngayinye yento inikela emandleni e-kinetic ajikelezayo ngokusekelwe kokubili kubunzima bayo kanye nebanga layo ukusuka ku-axis squared.
Isikhathi sokungaguquguquki kwenhlayiya sishintsha kanjani uma isisindo sayo siphindwe kathathu ngenkathi kugcinwa ibanga ukusuka ku-axis constant?
- Impendulo: Uma isisindo siphindwe kathathu futhi ibanga lihlala lingaguquki, isikhathi sokungagxili naso sizophindeka kathathu ngoba silingana ngqo nobunzima.
Ingabe inhlayiya ingaba nomzuzu wokungapheleli okungu-zero? Uma kunjalo, ngaphansi kwasiphi isimo?
- Impendulo: Yebo, inhlayiya izoba nomzuzu we-inertia ongu-zero uma itholakala ngqo ku-axis yokujikeleza, okwenza ibanga layo libe lide. $r$ kusukela ku-axis elingana no-zero.
Kungani izinto ezahlukene ezinobunzima nobukhulu obufanayo zinezikhathi ezahlukene zokungahlali kahle lapho zijikeleza ngezimbazo ezahlukene?
- Impendulo: Ukusatshalaliswa kwesisindo mayelana ne-axis yokujikeleza kunquma isikhathi se-inertia. Ngisho noma izinto ezimbili zinesisindo nosayizi ofanayo, ukusatshalaliswa kwazo kwesisindo maqondana ne-axis yokujikeleza kungahluka, okuholela ezikhathini ezahlukene ze-inertia.
Ingabe umzuzu we-inertia uyinani le-scalar noma le-vector?
- Impendulo: Isikhathi sokungagxili siyinani elilinganiselwe. Kodwa-ke, ngemizimba eqinile enezimo eziyinkimbinkimbi kanye nama-axes amaningi okujikeleza, imelelwa yi-tensor.
Uma izinhlayiya ezimbili, isisindo ngasinye $m$ , zitholakala kude $r_{1}$ futhi $r_{2}$ kusukela ku-axis yokujikeleza, iyini i-combined inertia moment?
- Impendulo: Umzuzu we-inertia uyisengezo sezinhlayiya ezihlukene. Ngakho-ke, umzuzu ohlangene we-inertia $I = m \times r_{12} + m \times r_{22}$ .
Isikhathi sokungagxili sihlobene kanjani nokulondolozwa komfutho we-angular?
- Impendulo: Umfutho we-angular $L$ kungumkhiqizo wesikhathi sokungahlali isikhathi eside $I$ kanye nesivinini se-angular $ω$ , emelelwe yi-equation $L = I \times ω$ Uma kungekho ma-torque angaphandle asebenza ohlelweni, umfutho we-angular uzohlala ungaguquki. Lokhu kusho ukuthi uma umzuzu we-inertia ushintsha (njengasekudonseni kwe-figure skater ezingalweni zakhe), ijubane le-angular kumele livumelane ukuze ligcine umkhiqizo ungaguquki.