Su'aalaha Tusaalaha ah ee Dhaqdhaqaaqa Wareega

11 Tusaalooyin Su'aalo ah oo ku saabsan Dhaqdhaqaaqa Wareega

Daqiiqadda Qaabka

1. Ul aad u fudud, dhererkeeduna yahay 140 cm. Saddex xoog ayaa ku shaqeeya usha, mid walbana wuxuu leeyahay cabbir F ah.₁ = 20 Newton, F₂ = 10 N, iyo F₃ = 40 N oo leh jihada iyo booska sida ku cad sawirka. Cabbirka daqiiqadda xoogga ee sababa in ushu ku wareegto bartamaha cufkeeda waa...

A. 40 Nm

B. 39 Nm

C. 28 Nm

D. 14 Nm

E. 3 Nm

Dood

Waa la ogyahay in:

Bartamaha cufka usha wuxuu ku yaal bartamaha usha.

Dhererka usha (l) = 140 cm = 1,4 mitir

Qaabka 1 (F)₁) = 20 N, gacanta xoogga 1 (l₁) = 70 cm = 0,7 mitir

Qaabka 2 (F)₂) = 10 N, gacanta xoogga 2 (l₂) = 100 cm – 70 cm = 30 cm = 0,3 mitir

Qaabka 3 (F)₃) = 40 N, gacanta xoogga 3 (l₃) = 70 cm = 0,7 mitir

La weydiiyay: Baaxadda daqiiqadda xoogga ee sababa in ushu ku wareegto bartamaha cufkeeda

Jawaab:

Daqiiqadda xoogga 1 waxay sababtaa in ushu u wareegto dhanka saacadda. Sidaa darteed, daqiiqadda xoogga 1 waa taban.

τ₁ =F₁ l₁ = (20 N)(0,7 m) = -14 N m

Daqiiqadda xoogga 2 waxay sababtaa in ushu ay u wareegto dhanka ka soo horjeeda saacadda. Sidaa darteed, daqiiqadda xoogga 2 waa mid togan.

τ₂ =F₂ l₂ = (10 N)(0,3 m) = 3 N m

Daqiiqadda xoogga 3 waxay sababtaa in ushu u wareegto dhanka saacadda. Sidaa darteed, daqiiqadda xoogga 3 waa taban.

τ₃ =F₃ l₃ = (40 N)(0,7 m) = -28 N m

Daqiiqad xoog leh oo ka dhalatay:

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

Cabbirka daqiiqadda xoogga waa 39 mitir oo Newton ah. Calaamadda taban waxay ka dhigan tahay in ushu ay u wareegto dhanka ka soo horjeeda saacadda.

Jawaabta saxda ah waa B.

2. Rod AB, oo cufkiisa la iska indho tiray, ayaa si toosan loo dhigay waxaana ku dhaqaaqay saddex xoog sida ku cad sawirka. Daqiiqad xoog leh oo ka dhalatay kaas oo ku shaqeeya usha marka lagu wareejiyo dhidibka D waa… (dembiga 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

Dood

Waa la garanayaa :

Dhidibka wareegga ama rogidda wuxuu ku yaal barta D.

F₁ = 10 N iyo l₁ =r₁ sin θ = (40 cm) (dembi 53^o) = (0,4 m)(0,8) = 0,32 mitir

F₂ = 10√2 N iyo l₂ =r₂ sin θ = (20 cm) (dembi 45^o) = (0,2 m)(0,5√2) = 0,1√2 mitir

F₃ = 20 N iyo l₃ =r₁ sin θ = (10 cm) (dembi 90^o) = (0,1 m)(1) = 0,1 mitir

La weydiiyay Daqiiqad xoog leh oo ka dhalatay

Jawab :

τ₁ =F₁ l₁ = (10 N)(0,32 m) = 3,2 Nm

(wanaagsan sababtoo ah daqiiqaddan xoogga ah waxay sababtaa in balooggu uu u wareego dhanka ka soo horjeeda saacadda)

τ₁ =F₂ l₂ = (10√2 N)( 0,1√2 m) = -2 Nm

(taban sababtoo ah daqiiqaddan xoogga ah waxay sababtaa in balooggu uu u wareego saacadda)

τ₁ =F₂ l₂ = (20 N)(0,1 m) = 2 Nm

(wanaagsan sababtoo ah daqiiqaddan xoogga ah waxay sababtaa in balooggu uu u wareego dhanka ka soo horjeeda saacadda)

Daqiiqad xoog leh oo ka dhalatay:

Στ = τ₁ – τ₁ + τ₃

Στ = 3,2 Nm – 2 Nm + 2 Nm

Στ = 3,2 Nm

Jawaabta saxda ah waa D.

3. Rod AB, oo cufkiisa la iska indho tiro, ayaa si toosan loo dhigay waxaana ku dhaqaaqay saddex xoog sida ku cad sawirka. Daqiiqadda ka dhalatay xoogga ku shaqeeya usha marka lagu wareejiyo dhidibka D waa… (dembiga 53^o = 0,8)

A. 2,4 Nm

B. 2,6 Nm

C. 3,0 Nm

D. 3,2 Nm

E. 3,4 Nm

Dood

Waa la garanayaa :

Dhidibka wareegga wuxuu ku yaal D.

Masaafada u dhaxaysa F₁ iyo dhidibka wareegga (r)_AD) = 40 cm = 0,4 m

Masaafada u dhaxaysa F₂ iyo dhidibka wareegga (r)_BD) = 20 cm = 0,2 m

Masaafada u dhaxaysa F₃ iyo dhidibka wareegga (r)_CD) = 10 cm = 0,1 m

F₁ = 10 Newton

F₂ = 10√2 Newton

F₃ = 20 Newton

Dembiga 53^o = 0,8

La weydiiyay Daqiiqo xoog ah oo ka dhalan karta haddii usha lagu wareejiyo dhidibka D

Jawab :

Xisaabi daqiiqadda xoogga uu soo saaro xoog kasta.

Daqiiqad xoog leh 1

Στ₁ = (F)₁)(r_AD aan lahayn 53^o) = (10 N)(0,4 m)(0,8) = 3,2 Nm

Daqiiqadda xoogga 1 waa mid togan sababtoo ah jihada wareegga usha oo ay keento daqiiqadda xoogga 1 waa mid ka soo horjeedda saacadda.

Daqiiqad xoog leh 2

Στ₂ = (F)₂)(r_BD aan lahayn 45^o) = (10√2 N)(0,2 m)(0,5√2) = -2 Nm

Daqiiqadda xoogga 2 waa taban sababtoo ah jihada wareegga usha oo ay keento daqiiqadda xoogga 2 waxay ku socotaa jihada la mid ah wareegga gacmaha saacadda.

Daqiiqad xoog leh 3

Στ₃ = (F)₃)(r_CD aan lahayn 90^o) = (20 N)(0,1 m)(1) = 2 Nm

Daqiiqadda xoogga 3 waa mid togan sababtoo ah jihada wareegga usha oo ay keento daqiiqadda xoogga 3 waa mid ka soo horjeedda saacadda.

Daqiiqad xoog leh oo ka dhalatay

Στ = Στ₁ + Στ₂ + Στ₃

Στ = 3,2 – 2 + 2

Στ = 3,2 mitir oo Newton ah

Jawaabta saxda ah waa D.

Daqiiqadda Inertia

4. Fiiri sawirka laba kubbadood oo ay silig isku xiran yihiin. Dhererka siliggu = 12 m, l₁ = 4 m oo cufka fiilada la iska indho tiro, markaas baaxadda daqiiqadda firfircoonida nidaamka waa…

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

Dood

Waa la garanayaa :

Cufka kubadda A (m)_A) = 0,2 kg

Cufka kubadda B (m)_B) = 0,6 kg

Masaafada u dhaxaysa kubadda A iyo dhidibka wareegga (r)_A) = 4 mitir

Masaafada u dhaxaysa kubadda B iyo dhidibka wareegga (r)_B) = 12 – 4 = 8 mitir

La weydiiyay : Daqiiqad firfircooni ah (I) nidaamka

Jawab :

Daqiiqad firfircoon oo kubadda A ah

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

Daqiiqadda firfircoonida kubadda B

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

Daqiiqadda firfircoonida nidaamka walxaha :

I = aniga_A + Aniga_B = 3,2 + 38,4 = 41,6 kg m²

Jawaabta saxda ah waa B.

Sharciga Labaad ee Newton ee Dhaqdhaqaaqa Wareegga

5. Ka fiirso sawirka giraangir adag oo isku mid ah oo hoos ku qoran. Xarig ayaa lagu duubay geeska giraangiraha, ka dibna dhammaadka xarigga waxaa lagu jiidayaa xoog F oo ah 6 N. Haddii cufka giraangiruhu yahay 5 kg oo gacantiisuna tahay 20 cm, dardargelinta xagasha giraangiraha waa…

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

Dood

Waa la ogyahay in:

Xoogga jilicsanaanta (F) = 6 Newton

Cufka taayirada (M) = 5 kg

Gacanka taayirada (R) = 20 cm = 20/100 m = 0,2 m

La weydiiyay: Dardargelinta xagasha giraangiraha (α)

Jawaab:

Xisaabi daqiiqadda xoogga:

τ = FR = (6 Newton)(0,2 mitir) = 1,2 mitir oo Newton ah

Xisaabi daqiiqadda firfircoonida:

Qaacidada loogu talagalay daqiiqadda firfircoonida giraangiraha qaabka saxanka adag waa 1/2 MR.² = 1/2 (5 kg)(0,2 m)² = 1/2 (5 kg)(0,04 m²) = 1/2 (0,2) = 0,1 kg m².

Xisaabi xawaaraha xagasha adoo isticmaalaya qaacidada dhaqdhaqaaqa wareega:

τ = waxaan ahay

α = τ / I = 1,2 / 0,1 = 12 rad s^-2

Jawaabta saxda ah waa E.

6. Barkin adag oo saxan ah oo miisaankiisu yahay 8 kg iyo gacan 10 cm ah ayaa lagu duubay geeska xarigga oo miisaankiisu yahay 4 kg oo ku xiran hal dhinac (g = 10 ms).^-2 ) Dardargelinta dhaqdhaqaaqa hoos u dhaca ee culayska waa...

A. 2,5 ms^-2

B. 5,0 ms^-2

C. 10,0 ms^-2

D. 20,0 ms^-2

Q. 33,3 ms^-2

Dood

Waa la ogyahay in:

Cufka saxanka adag (m) = 8 kg

Radius-ka saxanka adag (r) = 10 cm = 0,1 mitir

Cufka culayska (m) = 4 kg

Dardargelinta cufisjiidadka awgeed (g) = 10 m/s²

Miisaanka culayska (w) = mg = (4 kg)(10 m/s)²) = 40 kg m/s² = 40 Newton

La weydiiyay: Dardargelinta dhaqdhaqaaqa hoos u dhaca ee culayska

Jawaab:

Xisaabi daqiiqadda firfircoonida ee saxanka adag:

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

Xisaabi daqiiqadda xoogga:

τ = F r = (40 N) (0,1 m) = 4 Nm

Xisaabi xawaaraha xagasha adoo isticmaalaya qaacidada sharciga labaad ee Newton ee dhaqdhaqaaqa wareegga:

Στ = I α

4 = 0,04 α

α = 4 / 0,04 = 100

Xisaabi dardargelinta dhaqdhaqaaqa hoos u dhaca ee culayska:

a = r α = (0,1) (100) = 10 m/s²

Jawaabta saxda ah waa C.

7. Barkin adag oo leh cuf (M) iyo gacan (R) sida sawirka ka muuqata! Dhammaadka xarig aan tiro lahayn ayaa lagu duubay barkinta, dhammaadka kale ee xarigga waxaa lagu dhejiyay culays m kg ah, dardargelinta xagasha ee barkinta (α) haddii culaysku sii daayo. Haddii gabal balaastig ah oo A ah oo leh cufnaan 1⁄2 M ah lagu dhejiyo barkinta, si loo soo saaro Dardargelinta xagal isku mid ah ee culayska waa in la sameeyaa…. (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

Dood

Waa la garanayaa :

cufka culayska = m

Miisaanka culayska = w = mg

Cufka roogga adag = M

Radius of pulley adag = R

Dardargelinta xagasha ee pulley = α

La weydiiyay :

Haddii cufka jiidhku uu kordho M + M/2 = 3M/2 iyo dardargelinta xagasha jiidhku = α, waa maxay cufka culaysku?

Jawab :

Daqiiqadda firfircoonida ee khaanadaha aan lahayn balaastikada:

I = 1/2 MR² = 0,5 MR²

Daqiiqadda firfircoonida ee boolal + balaastiig:

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

Daqiiqad xoog leh:

τ = FR

Sharciga labaad ee dhaqdhaqaaqa wareegga ee Newton:

Στ = I α

w R = I α

mg R = I α

α = mg R / I

Si loo soo saaro dardargelin xagal isku mid ah, cufka culayska waa in la sameeyaa….. Ku beddel α ee isla'egta 2 iyo α ee isla'egta 1:

Jawaabta saxda ah waa B.

8 .. Shaxanka waxaa lagu muujiyay jiid adag oo xarig ku duuban geeskiisa dibadda. Iska hor imaadka u dhexeeya jiidka iyo xarigga iyo is jiidhka dhidibka wareegga waa la dayacay. Haddii culaysku hoos ugu dhaco xawaare joogto ah a ms^-2, markaa qiimaha daqiiqadda inertia ee pulley-ga wuxuu la mid yahay….

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

Dood

Waa la ogyahay in:

Xoog = w = mg

Gacan xoog leh = R

Dardargelinta xagasha = α

Dardargelinta rarka = a ms^-2

La weydiiyay: Daqiiqadda firfircoonida ee roogga (I)

Jawaab:

Xiriirka ka dhexeeya dardargelinta toosan iyo dardargelinta xagasha:

a = R α

α = a / R

Daqiiqada inertia waxaa lagu xisaabiyaa qaacidada:

τ = waxaan ahay

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

Jawaab sax ah ma jirto.

9. Barkin laga sameeyay walxo adag oo xarig lagu duubay dhinaciisa dibadda ayaa lagu muujiyay sida ka muuqata sawirka. Iska horimaadka barkinta waa la dayacay. Haddii daqiiqadda is-dhexgalka barkinta I = β iyo xarigga lagu jiido xoog joogto ah F, markaa qiimaha F wuxuu la mid yahay….

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

Dood

Waa la ogyahay in:

Xoogga jiidista = F

Daqiiqada firfircoonida ee roogga = β

Dardargelinta xagasha ee pulley = α

Gacanka jiidista = R

La weydiiyay: Qiimaha F wuxuu la mid yahay….

Jawaab:

Sharciga labaad ee dhaqdhaqaaqa wareegga ee Newton:

Στ = β α ———- Isla'egta 1

Sharaxaadda qaacidada:

Στ = Daqiiqad xoog ah oo natiijada ka dhalatay (torque)

β = Daqiiqada firfircoonida

α = Dardargelinta xagasha

Daqiiqad xoog ah oo ka dhalatay falgalka shabagga:

Στ = FR ———-> Isla'egta 2

Sharaxaadda qaacidada:

F = xoog jiidis ah

R = Masaafada laga bilaabo barta ficilka xoogga F ilaa dhidibka wareegga = gacanka shabagga

Ku beddel Στ isla'egta 1 Στ ee isla'egta 2:

Στ = β. α

F. R = β. α

F = (β.α) / R

F = β. α. (R^-1)

Jawaabta saxda ah waa D.

Dhaqdhaqaaqa Angular

10. Walax cufkeedu yahay 0,2 garaam ayaa ku wareegaysa goobada iyadoo xawaaraha xagal wali 10 rad s^-1Haddii gacanka wadada ay ku socoto qaybta uu yahay 3 cm, markaa dhaqdhaqaaqa xagasha ee qaybta waa...

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

Dood

Waa la ogyahay in:

Cufka walxaha (m) = 0,2 garaam = 2 x 10^-4 kg

Xawaaraha xagasha (ω) = 10 rad s^-1

Gacanka wadada walxaha (r) = 3 cm = 3 x 10^-2 meter

La weydiiyay: Dhaqdhaqaaq xagal ah oo walax ah

Jawaab:

Qaacidada dhaqdhaqaaqa xagasha:

L = I ω

Sharaxaad: I = dhaqdhaqaaqa xagal, I = daqiiqadda firfircoonida, ω = xawaaraha xagal

Daqiiqadda firfircoonida walxaha:

Aniga = Mudane² = (2 x 10^-4 )(3 x 10^-2)² = (2 x 10^-4 )(9 x 10^-4) = 18 x 10^-8

Dhaqdhaqaaqa xagasha waa:

L = I ω = (18 x 10^-8)(10 rad s^-1) = 18 x 10^-7 kg m² s^-1

Jawaab sax ah ma jirto.

11. Qoob-ka-cayaaruhu wuu wareegayaa iyadoo gacmaheeda la fidsan yahay ilaa 160 cm. Kadib, gacmaheeda waxaa la laabmaa ilaa 80 cm oo ku yaal xusullada. Haddii xawaaraha xagasha qoob-ka-cayaaruhu uu ahaado mid joogto ah, markaa dhaqdhaqaaqeeda toosan...

A. wali

B. wuxuu noqonayaa 1/2 cabirkii asalka ahaa

C. wuxuu noqonayaa 3/4 cabirkii asalka ahaa

D. wuxuu noqonayaa 2 jeer kii asalka ahaa

E. wuxuu noqonayaa 4 jeer oo asalka ah

Dood

Waa la ogyahay in:

Gacanka 1 (r)₁) = 160cm

Gacanka 2 (r)₂) = 80cm

Xawaaraha xagasha 1 (ω)₁) = ω

Xawaaraha xagasha 1 (ω)₂) = ω

La weydiiyay: Dhaqdhaqaaq toosan

Jawaab:

Xawaaraha toosan 1:

v₁ =r₁ ω₁ = (160 cm) ω

Xawaaraha toosan 2:

v₂ =r₂ ω₂ = (80 cm) ω

Dhaqdhaqaaqa toosan 1:

p = mv₁ = m (160 cm) ω

Dhaqdhaqaaqa toosan 2:

p = mv₂ = m (80 cm) ω

Markaa dhaqdhaqaaqa toosan wuxuu noqonayaa 1/2 jeer asalka.

Jawaabta saxda ah waa B.

Isha su'aasha:

Su'aalaha Imtixaanka Qaranka ee Fiisigiska ee Dugsiga Sare/Dugsiga Sare ee Xirfadda