Isikhathi se-inertia

1. Isikhathi sokungahlali kahle kwenhlayiya

Buyekeza inhlayiya ejikelezayo. Inhlayiya enesisindo m inikezwa amandla F ukuze inhlayiya ijikeleze i-axis O. Inhlayiya ihlukile ku-axis yokujikeleza. Okokuqala, inhlayiya iphumule (v = 0). Ngemva kokuhamba ngamandla ka-F, izinhlayiya zihamba ngesivinini esithile ukuze izinhlayiya zibe nokusheshisa okuqondile. Ubudlelwano phakathi kwamandla (F), isisindo (m), kanye nokusheshisa okuqondile kwezinhlayiya kuvezwa yi-equation 3:

F = ma _tan (Isibalo 3)

Izinhlayiya zijikeleza ukuze izinhlayiya zibe nokusheshisa kwe-angular. Ubudlelwano phakathi kokusheshisa kwe-tangential kanye nokusheshisa kwe-angular buvezwa yi-equation 4:

i _-tan = r α (Isibalo 4)

Faka u-a ku-equation 3 esikhundleni sika-a ku-equation 4:

F = mr α

r F = mr ² α

τ = (mr ² ) α (Isibalo 5)

r F umzuzu wamandla kanti mr ² umzuzu wokungapheleli kwenhlayiya. Isibalo 5 sisho ubudlelwano phakathi komzuzu wamandla, umzuzu wokungapheleli, kanye nokusheshiswa kwe-angular kwezinhlayiya ezijikelezayo. Isibalo 5 ukulingana komthetho wesibili kaNewton wezinhlayiya ezijikelezayo.

Isikhathi sokungapheleli kwenhlayiya siwumkhiqizo wobuningi benhlayiya (m) kanye nesikwele sebanga phakathi kwezinhlayiya ezine-axis yokujikeleza (r2 ⁾.

I = mr ² (Isibalo 6)

I = umzuzu wokungabi namandla kwenhlayiya, m = isisindo senhlayiya, r = ibanga phakathi kwenhlayiya kanye ne-axis yokujikeleza. Isibalo 6 sisetshenziselwa ukunquma umzuzu wokungabi namandla kwenhlayiya.

3.2 Izinkinga zesampula zesikhathi sokungabi namandla kwenhlayiya

Inkinga yesampula 1.

Inhlayiya enesisindo esingama-2 kg yayiboshelwe entanjeni engamamitha angu-0.5 ubude bese ijikeleziswa. Iyini isikhathi sokungahlali kahle kwezinhlayiya lapho zijikeleza?

Isixazululo:

Mina = uMnu. ²

I = (2 kg) (0.5 m) ²

I = 0.5 kg m ²

Inkinga yesampula 2.

Izinhlayiya ezimbili, ngayinye enesisindo esingu-2 kg kanye no-4 kg, zixhunywe ngentambo elula, lapho ubude bentambo bungamamitha amabili. Unganaki isisindo sezintambo. Thola isikhathi sokungabi namandla kwezinhlayiya ezimbili, uma:

a) I-axis yokujikeleza itholakala phakathi kwezinhlayiya ezimbili

Isixazululo:

I = m ₁ r ₁ ² + m ₂ r ₂ ²

I = (2)(12) + (4)(12)

I = 2 + 4

I = 6 kg m ²

b) I-axis yokujikeleza ikude ngamamitha angu-0.5 ukusuka ku-particle enesisindo esingu-2 kg

I = m ₁ r ₁ ² + m ₂ r ₂ ²

I = (2)(0.5 ² ) + (4)(1.5 ² )

I = (2)(0.25) + (4)(2.25)

I = 0.5 + 9

I = 9.5 kg m ²

c) I-axis yokujikeleza ikude ngamamitha angu-0.5 ukusuka ku-particle enesisindo esingu-4 kg

I = m ₁ r ₁ ² + m ₂ r ₂ ²

I = (2)(1.52) + (4)(0.5 ² )

I = (2)(2.25) + (4)(0.25)

I = 4.5 + 1

I = 5.5 kg m ²

Ngokusekelwe emiphumeleni yezibalo ezingenhla, umzuzu wokungalali uthonywa kakhulu indawo ye-axis yokujikeleza. Nakuba ukuma nosayizi kufana ngoba indawo ye-axis yokujikeleza ihlukile, umzuzu wokungalali nawo uhlukile.

Izinhlayiya eziseduze kwe-axis yokujikeleza zinomzuzu omncane we-inertia, kanti izinhlayiya ezikude ne-axis yokujikeleza zinomzuzu omkhulu we-inertia. Uma sicabanga ukuthi izinhlayiya ezimbili ezingenhla ziyizinto eziqinile, khona-ke inhlayiya ngayinye iseduze ne-axis yokujikeleza.

inesikhathi esincane sokungangeni kunomzuzu wokungangeni kwenhlayiya ekude kakhulu ne-axis yokujikeleza.

3.3 Isikhathi sokungahlali komzimba oqinile ofanayo

Izinto eziqinile zakhiwe yizinhlayiya eziningi. Isikhathi sokungagxili komzimba oqinile yinani eliphelele lezikhathi zokungagxili komzimba ngamunye owakha into.

I = Σ mr ²

Mina = m₁ r₁² +m₂ r₂² +m₃ r₃² + ….. + m_n r_n²

Ukuze sinqume isikhathi sokungapheleli kwento eqinile, sidinga ukubuyekeza into uma ijikeleza ngoba indawo ye-axis yokujikeleza ithinta isikhathi sokungapheleli kwento. Isikhathi sokungapheleli (I) senhlayiya ngayinye sincike nasebuningini (m) benhlayiya kanye nesikwele sebanga (r ² ) lenhlayiya kusukela ku-axis yokujikeleza. Ubunzima bazo zonke izinhlayiya ezakha into buyisisindo sento.

Inkinga iwukuthi, ibanga lenhlayiya ngayinye ukusuka ku-axis yokujikeleza liyahlukahluka.

Buyekeza ukwehla kwefomula yomzuzu we-inertia wendandatho encane eneminwe engu-R kanye nesisindo esingu-M.

Uma i-axis of rotation itholakala maphakathi nendandatho, zonke izinhlayiya zendandatho encane zitholakala kusukela ku-axis of rotation. Isikhathi sokungapheleli kwendandatho encane silingana nesikhathi sokungapheleli kwazo zonke izinhlayiya:

Mina = Σ mr²

Mina = m₁ r₁² +m₂ r₂² +m₃ r₃² + ….. + m_n r_n²

Ubunzima bazo zonke izinhlayiya = ubuningi bendandatho encane (M)

I = M (r ₁ ² + r ₂ ² + r ₃ ² + ….. + r _n ² )

Inhlayiya ngayinye yendandatho encane ingu-r kusukela ku-axis yokujikeleza ukuze u-r ₁ = r ₂ = r ₃ = rn = R

Isibalo somzuzu we-inertia wendandatho encane:

I = MR ²

I = umzuzu wokungabi namandla kwendandatho encane, M = isisindo sendandatho encane, R = irediyasi yendandatho encane.

Kuthiwani uma i-axis of rotation ingekho maphakathi nendandatho? Uma i-axis of rotation ingekho maphakathi nendandatho, ifomula ye-moment of inertia yendandatho encane ayikwazi ukutholwa kusetshenziswa indlela engenhla,

ngoba ibanga lenhlayiya ngayinye ukusuka ku-axis yokujikeleza lihlukile. Ukuthola ifomula yesikhathi sokungahlali kahle kwenkinga efana nale akuxoxwa ngayo kulesi sihloko.

Ifomula yesikhathi sokungahlali isikhathi eside kwezinye izinto eziqinile

3.4 Inkinga yesampula yesikhathi sokungabi namandla komzimba oqinile ofanayo

Inkinga yesampula 1:

Isilinda senduku i-AB sinobunzima obungama-3 kg uma sijikeleziswa ku-B,

isikhathi sokungakhathali singama-27 kg m ^2. Siyini isikhathi sokungakhathali uma

kujikelezwe ku-C?

Isixazululo:

Isixazululo:

Ifomula yesikhathi sokungahlali kahle kwenduku eqinile ehambisanayo (i-axis of rotation isenkabeni yenduku):

I = 1/12 ML ²

27 = (1/12) ML ²

(27)(12) = ML ²

324 = ML ²

Ifomula yomzuzu we-inertia wenduku eqinile ehambisanayo (i-axis yokujikeleza emaphethelweni enduku):

I = 1/3 ML ²

I = 1/3 (324)

I = 108 kg m ²