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, ngayinye inobukhulu obungu-F.₁ = 20 Newton, F₂ = 10 N, kanye no-F₃ = 40 N ngesiqondiso kanye nendawo njengoba kuboniswe esithombeni. Ubukhulu besikhathi samandla esibangela ukuthi 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

Isitayela 1 (F)₁) = 20 N, ingalo yokuphoqelela 1 (l₁) = 70 cm = 0,7 amamitha

Isitayela 2 (F)₂) = 10 N, ingalo yokuphoqelela 2 (l₂) = 100 cm – 70 cm = 30 cm = 0,3 amamitha

Isitayela 3 (F)₃) = 40 N, ingalo yokuphoqelela 3 (l₃) = 70 cm = 0,7 amamitha

Kubuziwe: 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.

τ₁ =F₁ l₁ = (20 N)(0,7 m) = -14 N m

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

τ₂ =F₂ l₂ = (10 N)(0,3 m) = 3 N m

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

τ₃ =F₃ l₃ = (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. URod AB, osisindo sakhe singanakwa, ubekwa ngokuvundlile futhi usetshenziswa amandla amathathu njengoba kuboniswe esithombeni. Umzuzu wamandla ophumelayo okusebenza endukwini uma kujikeleziswa ku-axis ku-D… (isono 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

I-Diketahui :

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

F₁ = 10 N kanye no-l₁ =r₁ isono θ = (40 cm)(isono 53^o) = (0,4 m)(0,8) = 0,32 amamitha

F₂ = 10√2 N kanye no-l₂ =r₂ isono θ = (20 cm)(isono 45^o) = (0,2 m)(0,5√2) = 0,1√2 amamitha

F₃ = 20 N kanye no-l₃ =r₁ isono θ = (10 cm)(isono 90^o) = (0,1 m)(1) = 0,1 amamitha

Kubuziwe : Umzuzu wamandla ophumelayo

Jawab :

τ₁ =F₁ l₁ = (10 N)(0,32 m) = 3,2 Nm

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

τ₁ =F₂ l₂ = (10√2 N)(0,1√2 m) = -2 Nm

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

τ₁ =F₂ l₂ = (20 N)(0,1 m) = 2 Nm

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

Umzuzu wamandla ophumela:

Στ = τ₁ – τ₁ + τ₃

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… (isono 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

I-Diketahui :

I-axis yokujikeleza itholakala ku-D.

Ibanga eliphakathi kuka-F₁ kanye ne-axis yokujikeleza (r)_AD) = 40 cm = 0,4 m

Ibanga eliphakathi kuka-F₂ kanye ne-axis yokujikeleza (r)_BD) = 20 cm = 0,2 m

Ibanga eliphakathi kuka-F₃ kanye ne-axis yokujikeleza (r)_CD) = 10 cm = 0,1 m

F₁ = 10 uNewton

F₂ = 10√2 uNewton

F₃ = 20 uNewton

Ngaphandle kwama-53^o = 0,8

Kubuziwe : Umzuzu wamandla ophumelayo uma induku ijikeleziswa ku-axis ku-D

Jawab :

Bala isikhathi samandla akhiqizwa yiwo wonke amandla.

Isikhathi samandla 1

Στ₁ = (F₁)(r_AD ngaphandle 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

Στ₂ = (F₂)(r_BD ngaphandle 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

Στ₃ = (F₃)(r_CD ngaphandle 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 = Στ₁ + Στ₂ + Στ₃

Στ = 3,2 – 2 + 2

Στ = 3,2 amamitha e-Newton

Impendulo efanele ngu-D.

Isikhathi Sokungabi Naso

4. Bheka isithombe samabhola amabili axhunywe yintambo. Ubude bentambo = 12 m, l₁ = 4 m futhi isisindo sentambo singanakwa, khona-ke ubukhulu besikhathi sokungabi namandla kwesistimu…

A. 52,6 kg m²

B. 41,6 kg m²

C. 34,6 kg m²

D. 22,4 kg m²

E. 20,4 kg m²

Ingxoxo

I-Diketahui :

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) = 4 amamitha

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

Kubuziwe : Isikhathi sokungalali kahle (I) uhlelo

Jawab :

Isikhathi sokungakhathali kwebhola A

I_A = (m_A)(r_A²) = (0,2)(4)² = (0,2)(16) = 3,2 kg m²

Isikhathi sokungakhathali kwebhola B

I_B = (m_B)(r_B²) = (0,6)(8)² = (0,6)(64) = 38,4 kg m²

Isikhathi sokungahlali kahle kwesistimu yezinhlayiya :

Mina = Mina_A + Mina_B = 3,2 + 38,4 = 41,6 kg m²

Impendulo efanele ingu-B.

Umthetho Wesibili KaNewton Wokunyakaza Okujikelezayo

5. Cabanga ngesithombe sesondo eliqinile elifanayo eliboniswe ngezansi. Intambo igoqwa emaphethelweni esondo, 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

Kubuziwe: Ukusheshisa kwesondo okugobile (α)

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 elinomumo wediski ingu-1/2 MR² = 1/2 (5 kg)(0,2 m)² = 1/2 (5 kg)(0,04 m²) = 1/2 (0,2) = 0,1 kg m².

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 komthwalo 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/s²

Isisindo somthwalo (w) = mg = (4 kg)(10 m/s²) = 40 kg m/s² = 40 uNewton

Kubuziwe: Ukusheshisa kokunyakaza komthwalo phansi

Impendulo:

Bala isikhathi sokungabi namandla kwediski eqinile:

I = 1/2 MR² = 1/2 (8 kg)(0,1 m)² = (4 kg)(0,01 m²) = 0,04 kg m²

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/s²

Impendulo efanele ngu-C.

7. I-pulley eqinile enesisindo (M) kanye ne-radius (R) njengoba kuboniswe esithombeni! Ukuphela kwentambo engenasisindo kugoqwe nge-pulley, okunye ukuphela kwentambo kulenga 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, ukukhiqiza ukusheshisa okufanayo kwe-angular komthwalo kumele kwenziwe…. (I pulley = 1/2 MR²)

A. 3/4 m kg

B. 3/2 m kg

C. 2 m kg

D. 3 m kg

E. 4 m kg

Ingxoxo

I-Diketahui :

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?

Jawab :

Isikhathi sokungakhathali kwe-pulley ngaphandle kwe-plasticine:

I = 1/2 MR² = 0,5 MR²

Isikhathi sokungabi namandla kwe-pulley + i-plasticine:

I = 1/2 (3M/2) R² = (3M/4) R² = 0,75M R²

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 eqinile enentambo eboshwe emaphethelweni ayo angaphandle iboniswe esithombeni. Ukungqubuzana phakathi kwe-pulley nentambo kanye nokungqubuzana ku-axis of rotation akunakwa. Uma umthwalo wehla ngokusheshisa okuqhubekayo, i-ms^-2, khona-ke inani lomzuzu wokungakhathali 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 umthwalo = a ms^-2

Kubuziwe: 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 ngezinto eziqinile 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

Kubuziwe: 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 ijubane le-angular namanje ama-rad ayi-10^-1Uma irediyasi yendlela yenhlayiya ingu-3 cm, khona-ke umfutho we-angular wenhlayiya...

A. 3 × 10^-7 kg m² s^-1

B. 9 × 10^-7 kg m² s^-1

C. 1,6 × 10^-6 kg m² s^-1

D. 1,8 × 10^-4 kg m² s^-1

E. 4,5 × 10^-3 kg m² 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 imitha

Kubuziwe: Umfutho we-angular wenhlayiya

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:

Mina = uMnu.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 m² 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= 160cm

Irediyasi 2 (r)₂= 80cm

Ijubane le-angular 1 (ω₁) = ω

Ijubane le-angular 1 (ω₂) = ω

Kubuziwe: Umfutho oqondile

Impendulo:

Ijubane eliqondile 1:

v₁ =r₁ ω₁ = (160 cm) ω

Ijubane eliqondile 2:

v₂ =r₂ ω₂ = (80 cm) ω

Umfutho oqondile 1:

p = mv₁ = m (160 cm) ω

Umfutho oqondile 2:

p = mv₂ = 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