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, F1 = 20 Newton, F2 = 10 N, iyo F3 = 40 N, mid walbana wuxuu leeyahay jihada iyo booska sida ku cad sawirka. Cabbirka daqiiqadda xoogga ee sababa usha inay 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
Xoogga 1 (F 1 ) = 20 N, gacanta xoogga 1 (l 1 ) = 70 cm = 0,7 mitir
Xoogga 2 (F 2 ) = 10 N, gacanta xoogga 2 (l 2 ) = 100 cm – 70 cm = 30 cm = 0,3 mitir
Xoogga 3 (F 3 ) = 40 N, gacanta xoogga 3 (l 3 ) = 70 cm = 0,7 mitir
Su'aal: Baaxadda daqiiqadda xoogga ee keenta 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.
τ 1 = F 1 l 1 = (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.
τ 2 = F 2 l 2 = (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.
τ 3 = F 3 l 3 = (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 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 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 ogyahay in :
Dhidibka wareegga ama rogidda wuxuu ku yaal barta D.
F 1 = 10 N iyo l 1 = r 1 dembi θ = (40 cm) (sinka 53 o ) = (0,4 m) (0,8) = 0,32 mitir
F 2 = 10√2 N iyo l 2 = r 2 dembi θ = (20 cm) ( dembi 45 o ) = (0,2 m)(0,5√2) = 0,1√2 mitir
F 3 = 20 N iyo l 3 = r 1 dembi θ = (10 cm) (sinka 90 o ) = (0,1 m) (1) = 0,1 mitir
La weydiiyay : Daqiiqad xoog ah oo ka dhalatay
Jawaab :
τ 1 = F 1 l 1 = (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)
τ 1 = F 2 l 2 = (10√2 N) ( 0,1√2 m) = -2 Nm
(taban sababtoo ah daqiiqaddan xoogga ah waxay sababtaa in balooggu uu u wareego saacadda)
τ 1 = F 2 l 2 = (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 = τ 1 – τ 1 + τ 3
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 ogyahay in :
Dhidibka wareegga wuxuu ku yaal D.
Masaafada u dhaxaysa F 1 iyo dhidibka wareegga (r AD ) = 40 cm = 0,4 m
Masaafada u dhaxaysa F 2 iyo dhidibka wareegga (r BD ) = 20 cm = 0,2 m
Masaafada u dhaxaysa F 3 iyo dhidibka wareegga (r CD ) = 10 cm = 0,1 m
F1 = 10 Newtons
F 2 = 10√2 Newton
F3 = 20 Newtons
Dem 53 o = 0,8
Su'aal : Daqiiqad xoog leh oo ka dhalan karta haddii usha lagu wareejiyo dhidibka D
Jawaab :
Xisaabi daqiiqadda xoogga uu soo saaro xoog kasta.
Daqiiqad xoog leh 1
Στ 1 = (F 1 ) (r AD sin 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
Στ 2 = (F 2 )(r BD sin 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
Στ 3 = (F 3 )(r CD sin 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 = Στ 1 + Στ 2 + Στ 3
Στ = 3,2 – 2 + 2
Στ = 3,2 mitir oo Newton ah
Jawaabta saxda ah waa D.
Daqiiqadda Inertia
4. Ka fiirso sawirka laba kubbadood oo ay silig isku xiran yihiin. Dhererka siliggu = 12 m, l 1 = 4 m cufka siligguna waa la iska indha tiraa, sidaa darteed baaxadda daqiiqadda firfircoonida nidaamka waa…
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
Dood
Waa la ogyahay in :
Cufka kubadda A (m A ) = 0,2 kg
Cufka kubadda B (m B ) = 0,6 kg
Masaafada u dhaxaysa kubbadda A iyo dhidibka wareegga (r A ) = 4 mitir
Masaafada u dhaxaysa kubbadda B iyo dhidibka wareegga (r B ) = 12 – 4 = 8 mitir
Su'aal : Daqiiqada firfircoonida (I) ee nidaamka
Jawaab :
Daqiiqad firfircoon oo kubadda A ah
I A = (m A )(r A 2 ) = (0,2) (4) 2 = (0,2) (16) = 3,2 kg m 2
Daqiiqadda firfircoonida kubadda B
I B = (m B )(r B 2 ) = (0,6)(8) 2 = (0,6)(64) = 38,4 kg m 2
Daqiiqadda firfircoonida nidaamka walxaha :
I = I A + I B = 3,2 + 38,4 = 41,6 kg m 2
Jawaabta saxda ah waa B.
Sharciga Labaad ee Newton ee Dhaqdhaqaaqa Wareegga
5. Fiiri sawirka giraangir adag oo isku mid ah oo dhinaca ku yaal. 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
Su'aal: 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 daqiiqadda firfircoonida giraangiraha adag ee qaabka saxanka ama saxanka waa 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.
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 miisaanka waa...
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
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/s 2
Miisaanka culayska (w) = mg = (4 kg)(10 m/s 2 ) = 40 kg m/s 2 = 40 Newtons
Su'aal: Dardargelinta dhaqdhaqaaqa hoos u dhaca ee culayska
Jawaab:
Xisaabi daqiiqadda firfircoonida ee saxanka adag:
I = 1/2 MR 2 = 1/2 (8 kg)(0,1 m) 2 = (4 kg)(0,01 m 2 ) = 0,04 kg m 2
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/s 2
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 cuf m kg ah, dardargelinta xagasha ee barkinta (α) haddii cufku sii daayo. Haddii gabal balaastig ah oo A ah oo leh cuf 1⁄2 M ah lagu dhejiyo barkinta, si loo soo saaro dardargelin xagal oo isku mid ah waa in culayska la sameeyaa…. (I barkin = 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
Dood
Waa la ogyahay in :
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?
Jawaab :
Daqiiqadda firfircoonida ee khaanadaha aan lahayn balaastikada:
I = 1/2 MR 2 = 0,5 MR 2
Daqiiqadda firfircoonida ee boolal + balaastiig:
I = 1/2 (3M/2) R 2 = (3M/4) R 2 = 0,75M R 2
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.. Barkin laga sameeyay shay adag oo xarig ku duuban yahay dhinaciisa dibadda ayaa lagu muujiyay sida ku cad sawirka. Isjiidjiidka barkinta ee xarigga iyo isjiidjiidka dhidibkiisa wareega waa la dayacay. Haddii culaysku hoos ugu dhaco xawaare joogto ah a ms -2 , markaa qiimaha daqiiqadda isjiidjiidka ee barkinta waxay la mid tahay….
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
Su'aal: Daqiiqada 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 shay adag oo xadhig ku duuban yahay dhinaciisa dibadda ayaa lagu muujiyay sida ku cad 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
Su'aal: 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 socota goobaabin leh xawaare xagal oo joogto ah oo ah 10 rad s -1 . Haddii gacanka wadada walaxda uu yahay 3 cm, markaa dhaqdhaqaaqa xagasha walaxda waa...
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
Dood
Waa la ogyahay in:
Cufka walxaha (m) = 0,2 garaam = 2 x 10 -4 kg
Xawaaraha xagasha (ω) = 10 rad s -1
Radius-ka waddada walxaha (r) = 3 cm = 3 x 10 -2 mitir
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:
I = mr 2 = (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 m 2 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 waa...
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 ) = 160 cm
Gacanka 2 (r 2 ) = 80 cm
Xawaaraha xagasha 1 (ω 1 ) = ω
Xawaaraha xagasha 1 (ω 2 ) = ω
La weydiiyay: Dhaqdhaqaaq toosan
Jawaab:
Xawaaraha toosan 1:
v 1 = r 1 ω 1 = (160 cm) ω
Xawaaraha toosan 2:
v 2 = r 2 ω 2 = (80 cm) ω
Dhaqdhaqaaqa toosan 1:
p = mv 1 = m (160 cm) ω
Dhaqdhaqaaqa toosan 2:
p = mv 2 = 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
