Imibuzo Yesibonelo Se-Rotational Dynamics

Izibonelo eziyi-11 zemibuzo ye-Rotational Dynamics

Isikhathi Sesitayela

1. Induku elula kakhulu, ubude obungu-140 cm. Amandla amathathu asebenza enduku, F1 ₌ 20 Newton, _F2 = 10 N, kanye F3 ₌ 40 N, ngalinye linesiqondiso nendawo njengoba kuboniswe esithombeni. Ubukhulu besikhathi samandla esibangela ukuba induku ijikeleze phakathi nendawo yayo yobunzima...

A. 40 Nm

B. 39 Nm

C. 28 Nm

D. 14 Nm

E. 3 Nm

Ingxoxo

Kuyaziwa ukuthi:

Isikhungo sesisindo senduku siphakathi kwenduku.

Ubude benduku (l) = 140 cm = 1,4 amamitha

Amandla 1 (F ₁ ) = 20 N, ingalo yamandla 1 (l ₁ ) = 70 cm = 0,7 amamitha

Amandla 2 (F ₂ ) = 10 N, ingalo yamandla 2 (l ₂ ) = 100 cm – 70 cm = 30 cm = 0,3 amamitha

Amandla 3 (F ₃ ) = 40 N, ingalo yamandla 3 (l ₃ ) = 70 cm = 0,7 amamitha

Umbuzo: Ubukhulu besikhathi samandla esibangela ukuthi induku ijikeleze phakathi nendawo yayo yobunzima

Impendulo:

Umzuzu wamandla 1 ubangela ukuthi induku ijikeleze ngokwewashi. Ngakho-ke, umzuzu wamandla 1 awulungile.

τ ₁ = F ₁ l ₁ = (20 N)(0,7 m) = -14 N m

Umzuzu wamandla 2 ubangela ukuthi induku ijikeleze ngokuphambene newashi. Ngakho-ke, umzuzu wamandla 2 ulungile.

τ ₂ = F ₂ l ₂ = (10 N)(0,3 m) = 3 N m

Umzuzu wamandla 3 ubangela ukuthi induku ijikeleze ngokwewashi. Ngakho-ke, umzuzu wamandla 3 awulungile.

τ ₃ = F ₃ l ₃ = (40 N)(0,7 m) = -28 N m

Umzuzu wamandla ophumela:

Στ = -14 Nm + 3 Nm – 28 Nm = – 42 Nm + 3 Nm = -39 Nm

Ubukhulu bamandla angu-39 Newton metres. Uphawu olunegethivu lusho ukuthi induku ijikeleza ngokuphambene newashi.

Impendulo efanele ingu-B.

2. I-Rod AB, enesisindo esinganakwa, ibekwa ngokuvundlile futhi isetshenziswa amandla amathathu njengoba kuboniswe esithombeni. Umzuzu ophumelayo wamandla asebenza endukwini lapho ijikeleziswa ku-axis ku-D… (sin 53 ^o = 0,8)

A. 2,4 N m

B. 2,6 N m

C. 3,0 N m

D. 3,2 N m

E. 3,4 N m

Ingxoxo

Kuyaziwa ukuthi :

I-axis yokujikeleza noma ye-pivot itholakala endaweni D.

F ₁ = 10 N kanye no-l ₁ = r ₁ isono θ = (40 cm)(isono 53 ^o ) = (0,4 m)(0,8) = 0,32 imitha

F ₂ = 10√2 N kanye no-l ₂ = r ₂ isono θ = (20 cm)(isono 45 ^o ) = (0,2 m)(0,5√2) = 0,1√2 amamitha

F ₃ = 20 N kanye no-l ₃ = r ₁ isono θ = (10 cm)(isono 90 ^o ) = (0,1 m)(1) = 0,1 imitha

Kubuziwe : Isikhathi samandla esiphumelayo

Impendulo :

τ ₁ = F ₁ l ₁ = (10 N)(0,32 m) = 3,2 Nm

(kuhle ngoba lo mzuzu wamandla ubangela ukuthi ibhloko lijikeleze ngokuphambene newashi)

τ ₁ = F ₂ l ₂ = (10√2 N)( 0,1√2 m) = -2 Nm

(okubi ngoba lo mzuzu wamandla ubangela ukuthi ibhloko lijikeleze ngokwewashi)

τ ₁ = F ₂ l ₂ = (20 N)(0,1 m) = 2 Nm

(kuhle ngoba lo mzuzu wamandla ubangela ukuthi ibhloko lijikeleze ngokuphambene newashi)

Umzuzu wamandla ophumela:

Στ = τ ₁ – τ ₁ + τ ₃

Στ = 3,2 Nm – 2 Nm + 2 Nm

Στ = 3,2 Nm

Impendulo efanele ngu-D.

3. I-Rod AB, enesisindo esinganakwa, ibekwa ngokuvundlile futhi isetshenziswa amandla amathathu njengoba kuboniswe esithombeni. Umzuzu ophumelayo wamandla asebenza endukwini lapho ejikeleziswa ku-axis ku-D… (sin 53 ^o = 0,8)

A. 2,4 Nm

B. 2,6 Nm

C. 3,0 Nm

D. 3,2 Nm

E. 3,4 Nm

Ingxoxo

Kuyaziwa ukuthi :

I-axis yokujikeleza itholakala ku-D.

Ibanga eliphakathi kuka-F ₁ kanye ne-axis yokujikeleza (r _AD ) = 40 cm = 0,4 m

Ibanga eliphakathi kuka-F ₂ kanye ne-axis yokujikeleza (r _BD ) = 20 cm = 0,2 m

Ibanga eliphakathi kuka-F ₃ kanye ne-axis yokujikeleza (r _CD ) = 10 cm = 0,1 m

F ₁ = 10 Newton

F ₂ = 10√2 Newton

F ₃ = 20 Newton

Isono 53 ^o = 0,8

Umbuzo : Isikhathi samandla esiphumelayo uma induku ijikeleziswa ku-axis ku-D

Impendulo :

Bala isikhathi samandla akhiqizwa yiwo wonke amandla.

Isikhathi samandla 1

Στ ₁ = (F ₁ )(r _AD isono 53 ^o ) = (10 N)(0,4 m)(0,8) = 3,2 Nm

Isikhathi samandla 1 sihle ngoba indlela yokujikeleza kwenduku ebangelwa yisikhathi samandla 1 iphambene newashi.

Isikhathi samandla 2

Στ ₂ = (F ₂ )(r _BD isono 45 ^o ) = (10√2 N)(0,2 m)(0,5√2) = -2 Nm

Isikhathi samandla 2 asilungile ngoba isiqondiso sokuzungeza kwenduku esibangelwa umzuzu wamandla 2 sisehlangothini olufanayo nokujikeleza kwezandla zewashi.

Isikhathi samandla 3

Στ ₃ = (F ₃ )(r _CD isono 90 ^o ) = (20 N)(0,1 m)(1) = 2 Nm

Isikhathi samandla 3 sihle ngoba indlela yokujikeleza kwenduku ebangelwa yisikhathi samandla 3 iphambene newashi.

Umzuzu wamandla ophumelayo

Στ = Στ ₁ + Στ ₂ + Στ ₃

Στ = 3,2 – 2 + 2

Στ = 3,2 amamitha e-Newton

Impendulo efanele ngu-D.

Isikhathi Sokungabi Naso

4. Cabanga ngesithombe samabhola amabili axhunywe yintambo. Ubude bentambo = 12 m, l ₁ = 4 m kanye nobunzima bentambo abunakwa, ngakho ubukhulu besikhathi sokungapheleli kwesistimu…

A. 52,6 kg m²

B. 41,6 kg m ²

C. 34,6 kg m ²

D. 22,4 kg m ²

E. 20,4 kg m ²

Ingxoxo

Kuyaziwa ukuthi :

Isisindo sebhola A (m _A ) = 0,2 kg

Isisindo sebhola B (m _B ) = 0,6 kg

Ibanga eliphakathi kwebhola A kanye ne-axis yokujikeleza (r _A ) = amamitha angu-4

Ibanga eliphakathi kwebhola B kanye ne-axis yokujikeleza (r _B ) = 12 – 4 = 8 amamitha

Umbuzo : Isikhathi sokungakhathali (I) kwesistimu

Impendulo :

Isikhathi sokungakhathali kwebhola A

I _A = (m _A )(r _A ² ) = (0,2)(4) ² = (0,2)(16) = 3,2 kg m ²

Isikhathi sokungakhathali kwebhola B

I _B = (m _B )(r _B ² ) = (0,6)(8) ² = (0,6)(64) = 38,4 kg m ²

Isikhathi sokungapheleli kohlelo lwezinhlayiya :

Mina = I _A + I _B = 3,2 + 38,4 = 41,6 kg m ²

Impendulo efanele ingu-B.

Umthetho Wesibili KaNewton Wokunyakaza Okujikelezayo

5. Bheka isithombe sesondo eliqinile elifanayo ohlangothini. Intambo igoqwa onqenqemeni lwesondo bese kuthi ukuphela kwentambo kudonswe ngamandla F ka-6 N. Uma isisindo sesondo singama-5 kg ​​kanti i-radius yalo ingu-20 cm, ukusheshisa kwesondo okungu-angular kungu…

A. 0,12 rad s^-2

B. 1,2 rad s ^–2

C. 3,0 rad s ^–2

D. 6,0 rad s ^–2

E. 12,0 rad s ^–2

Ingxoxo

Kuyaziwa ukuthi:

Amandla okudonsa (F) = 6 Newton

Isisindo samasondo (M) = 5 kg

Irediyasi yesondo (R) = 20 cm = 20/100 m = 0,2 m

Umbuzo: Ukusheshisa kwesondo (α) nge-angular

Impendulo:

Bala isikhathi samandla:

τ = FR = (6 Newton)(0,2 amamitha) = 1,2 amamitha Newton

Bala isikhathi sokungabi naso isikhathi:

Ifomula yesikhathi sokungabi namandla kwesondo eliqinile ngesimo sediski noma ipuleti ingu-1/2 MR ² = 1/2 (5 kg)(0,2 m) ² = 1/2 (5 kg)(0,04 m ² ) = 1/2 (0,2) = 0,1 kg m ².

Bala ukusheshisa kwe-angular usebenzisa ifomula ye-rotation dynamics:

τ = ngi α

α = τ / I = 1,2 / 0,1 = 12 rad s ^-2

Impendulo efanele ngu-E.

6. I-pulley yediski eqinile enesisindo esingu-8 kg kanye ne-radius engu-10 cm igoqwe onqenqemeni lwentambo enesisindo esingu-4 kg esiboshelwe kolunye uhlangothi (g = 10 ms ^-2 ). Ukusheshisa kokunyakaza kwesisindo phansi...

A. 2,5 ms ^–2

B. 5,0 ms ^–2

C. 10,0 ms ^–2

D. 20,0 ms ^–2

E. 33,3 ms ^–2

Ingxoxo

Kuyaziwa ukuthi:

Isisindo se-pulley yediski eqinile (m) = 8 kg

Irediyasi yepulley yediski eqinile (r) = 10 cm = 0,1 amamitha

Isisindo somthwalo (m) = 4 kg

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Isisindo somthwalo (w) = mg = (4 kg)(10 m/s ² ) = 40 kg m/s ² = 40 Newtons

Umbuzo: Ukusheshisa ukunyakaza komthwalo phansi

Impendulo:

Bala isikhathi sokungabi namandla kwediski eqinile:

I = 1/2 MR ² = 1/2 (8 kg)(0,1 m) ² = (4 kg)(0,01 m ² ) = 0,04 kg m ²

Bala isikhathi samandla:

τ = F r = (40 N)(0,1 m) = 4 Nm

Bala ukusheshisa kwe-angular usebenzisa ifomula yomthetho wesibili kaNewton yokunyakaza okujikelezayo:

Στ = I α

4 = 0,04 α

α = 4 / 0,04 = 100

Bala ukusheshisa kokunyakaza komthwalo phansi:

a = r α = (0,1)(100) = 10 m/s ²

Impendulo efanele ngu-C.

7. I-pulley eqinile enesisindo (M) kanye ne-radius (R) njengoba kuboniswe esithombeni! Ukuphela kwentambo engenasisindo kugoqwe nge-pulley, olunye uhlangothi lwentambo lulengiswe ngomthwalo we-m kg, ukusheshisa kwe-angular kwe-pulley (α) uma umthwalo ukhishwa. Uma ucezu lwe-plasticine A olunesisindo esingu-1⁄2 M lunamathiselwe ku-pulley, ukuze kukhiqizwe ukusheshisa okufanayo kwe-angular umthwalo kumele wenziwe…. (I-pulley = 1/2 MR ² )

A. 3/4 m kg

B. 3/2 m kg

C. 2 m kg

D. 3 m kg

E. 4 m kg

Ingxoxo

Kuyaziwa ukuthi :

isisindo somthwalo = m

Isisindo somthwalo = w = mg

Isisindo se-pulley eqinile = M

Irediyasi ye-pulley eqinile = R

Ukusheshisa kwe-angular kwe-pulley = α

Kubuziwe :

Uma isisindo se-pulley sikhuphuka sibe yi-M + M/2 = 3M/2 kanye nokusheshisa kwe-angular kwe-pulley = α, siyini isisindo somthwalo?

Impendulo :

Isikhathi sokungakhathali kwe-pulley ngaphandle kwe-plasticine:

I = 1/2 MR ² = 0,5 MR ²

Isikhathi sokungabi namandla kwe-pulley + i-plasticine:

Mina = 1/2 (3M/2) R ² = (3M/4) R ² = 0,75M R ²

Isikhathi samandla:

τ = FR

Umthetho wesibili kaNewton wokunyakaza kokujikeleza:

Στ = I α

w R = I α

mg R = I α

α = mg R / I

Ukuze kukhiqizwe ukusheshisa okufanayo kwe-angular, isisindo somthwalo kumele senziwe….. Faka u-α esilinganisweni sesi-2 no-α esilinganisweni soku-1:

Impendulo efanele ingu-B.

8.. I-pulley eyenziwe ngento eqinile enentambo egoqwe ohlangothini lwayo lwangaphandle iboniswa njengoba kuboniswe esithombeni. Ukungqubuzana kwe-pulley nentambo kanye nokungqubuzana ku-axis yayo yokujikeleza kuyanganakwa. Uma umthwalo wehla ngokusheshisa okuqhubekayo okungu-ms ^-2 , khona-ke inani lomzuzu wokungagxili kwe-pulley lilingana no….

A. I = τ α R

B. I = τ α ^-1 R

C. I = τ a R

D. I = τ a ^-1 R ^-1

E. I = τ a R ^-1

Ingxoxo

Kuyaziwa ukuthi:

Amandla = w = mg

Ingalo yamandla = R

Ukusheshisa kwe-angular = α

Ukusheshisa komthwalo = a ms ^-2

Umbuzo: Isikhathi sokungahlali kahle kwe-pulley (I)

Impendulo:

Ubudlelwano phakathi kokusheshisa okuqondile kanye nokusheshisa kwe-angular:

a = R α

α = a/R

Isikhathi se-inertia sibalwa kusetshenziswa ifomula:

τ = ngi α

I = τ : α = τ : a / R = τ (R / a) = τ R a ^-1

Ayikho impendulo efanele.

9. I-pulley eyenziwe ngento eqinile enentambo eboshwe ohlangothini lwayo lwangaphandle iboniswa njengoba kuboniswe esithombeni. Ukungqubuzana kwe-pulley kunganakwa. Uma umzuzu wokungapheleli kwe-pulley I = β futhi intambo idonswa ngamandla angaguquki F, khona-ke inani lika-F lilingana no….

A. F = α .β .R

B. F = α . β ² . R

C. F = α . (β . R) ^-1

D. F = α . β . (R) ^-1

E. F = R. (α. β) ^-1

Ingxoxo

Kuyaziwa ukuthi:

Amandla okudonsa = F

Isikhathi sokungakhathali kwe-pulley = β

Ukusheshisa kwe-angular kwe-pulley = α

Irediyasi ye-Pulley = R

Umbuzo: Inani lika-F lilingana no….

Impendulo:

Ifomula yomthetho wesibili kaNewton yokunyakaza kokujikeleza:

Στ = β α ———- Isibalo 1

Incazelo yefomula:

Στ = Umphumela wamandla (torque)

β = Isikhathi sokungahlali isikhathi eside

α = Ukusheshisa kwe-Angular

Umzuzu wamandla ophumelayo osebenza ku-pulley:

Στ = FR ———-> Isibalo 2

Incazelo yefomula:

F = amandla okudonsa

R = Ibanga ukusuka endaweni yokusebenza kwamandla F kuya ku-axis yokujikeleza = irediyasi ye-pulley

Shintshanisa u-Στ esibalweni 1 ufake u-Στ esibalweni 2:

Στ = β . α

F. R = β. α

F = (β . α) / R

F = β . α . (R ^-1 )

Impendulo efanele ngu-D.

Umfutho We-Angular

10. Inhlayiya enesisindo esingamagremu angu-0,2 ihamba ngendilinga enejubane elingaguquguquki elingu -10 rad s ^-1 . Uma irediyasi yendlela yenhlayiya ingu-3 cm, khona-ke umfutho we-angle wenhlayiya...

A. 3 × 10 ^–7 kg m ² s ^-1

B. 9 × 10 ^–7 kg m ² s ^-1

C. 1,6 × 10 ^–6 kg m ² s ^-1

D. 1,8 × 10 ^–4 kg m ² s ^-1

E. 4,5 × 10 ^–3 kg m ² s ^-1

Ingxoxo

Kuyaziwa ukuthi:

Isisindo sezinhlayiya (m) = 0,2 amagremu = 2 x 10 ^-4 kg

Ijubane le-angular (ω) = 10 rad s ^-1

Irediyasi yendlela yezinhlayiya (r) = 3 cm = 3 x 10 ^-2 amamitha

Kubuziwe: Umfutho we-angular wezinhlayiya

Impendulo:

Ifomula yomfutho we-angular:

L = I ω

Incazelo: I = umfutho we-angular, I = umzuzu we-inertia, ω = ijubane le-angular

Isikhathi sokungahlali kahle kwezinhlayiya:

I = mr ² = (2 x 10 ^-4 )(3 x 10 ^-2 ) ² = (2 x 10 ^-4 )(9 x 10 ^-4 ) = 18 x 10 ^-8

Umfutho we-angular uthi:

L = I ω = (18 x 10 ^-8 )(10 rad s ^-1 ) = 18 x 10 ^-7 kg m ² s ^-1

Ayikho impendulo efanele.

11. Umdansi uyajika izingalo zakhe zeluliwe zibe ngu-160 cm ubude. Ngemuva kwalokho, izingalo zakhe zigoqwa zibe ngu-80 cm ubude ezindololwaneni. Uma ijubane lomdansi lihlala lingaguquguquki, khona-ke umfutho wakhe oqondile...

A. namanje

B. iba yi-1/2 yosayizi wokuqala

U-C. uba ngu-3/4 wosayizi wokuqala

U-D. uba ngokuphindwe kabili kunowokuqala

I-E. iba ngokuphindwe kane kuneyokuqala

Ingxoxo

Kuyaziwa ukuthi:

Irediyasi 1 (r ₁ ) = 160 cm

Irediyasi 2 (r ₂ ) = 80 cm

Ijubane le-angular 1 (ω ₁ ) = ω

Ijubane le-angular 1 (ω ₂ ) = ω

Kubuziwe: Umfutho oqondile

Impendulo:

Ijubane eliqondile 1:

v ₁ = r ₁ ω ₁ = (160 cm) ω

Ijubane eliqondile 2:

v ₂ = r ₂ ω ₂ = (80 cm) ω

Umfutho oqondile 1:

p = mv ₁ = m (160 cm) ω

Umfutho oqondile 2:

p = mv ₂ = m (80 cm) ω

Ngakho-ke umfutho oqondile uba ngokuphindwe ka-1/2 kunowokuqala.

Impendulo efanele ingu-B.

Umthombo wombuzo:

Imibuzo Yefiziksi Yezivivinyo Zikazwelonke Zesikole Samabanga Aphezulu/Isikole Samabanga Aphezulu Sokufundela Umsebenzi