Misalai 11 na Tambayoyin Juyawa
Lokacin Salo
1. Sanda mai sauƙi sosai, tsawon santimita 140. Ƙarfi uku suna aiki akan sandar, F1 = 20 Newton, F2 = 10 N, da F3 = 40 N, kowannensu yana da alkibla da matsayi kamar yadda aka nuna a cikin hoton. Girman lokacin ƙarfin da ke sa sandar ta juya a tsakiyar nauyinta shine...
A. 40 Nm
B. 39 Nm
C. 28 Nm
D. 14 Nm
E. 3 Nm
Tattaunawa
An san cewa:
Tsakiyar nauyin sandar tana tsakiyar sandar.
Tsawon sandar (l) = 140 cm = mita 1,4
Ƙarfi 1 (F1 ) = 20 N, hannun ƙarfi 1 (l1 ) = 70 cm = mita 0,7
Ƙarfi 2 (F 2 ) = 10 N, hannun ƙarfi 2 (l 2 ) = 100 cm – 70 cm = 30 cm = mita 0,3
Ƙarfi 3 (F3 ) = 40 N, hannun ƙarfi 3 (l3 ) = 70 cm = mita 0,7
Tambaya: Girman lokacin ƙarfin da ke sa sandar ta juya a tsakiyar nauyinta
Amsa:
Motsin ƙarfi 1 yana sa sandar ta juya a hannun agogo. Saboda haka, motsin ƙarfi 1 yana da korau.
τ 1 = F 1 l 1 = (20 N)(0,7 m) = -14 N m
Motsin ƙarfi na 2 yana sa sandar ta juya akasin agogo. Saboda haka, motsin ƙarfi na 2 yana da kyau.
τ 2 = F 2 l 2 = (10 N)(0,3 m) = 3 N m
Motsin ƙarfi 3 yana sa sandar ta juya a hannun agogo. Saboda haka, motsin ƙarfi 3 yana da korau.
τ 3 = F 3 l 3 = (40 N)(0,7 m) = -28 N m
Lokacin ƙarfi da ya biyo baya:
Στ = -14 Nm + 3 Nm - 28 Nm = - 42 Nm + 3 Nm = -39 Nm
Girman lokacin ƙarfin shine mita 39 na Newton. Alamar mara kyau tana nufin sandar tana juyawa akasin agogo.
Amsar da ta dace ita ce B.
2. Rod AB, wanda aka yi watsi da nauyinsa, an sanya shi a kwance kuma ƙarfi uku ne ke aiki da shi kamar yadda aka nuna a cikin hoton. Lokacin da ƙarfin da ke aiki akan sandar lokacin da aka juya shi akan axis a D shine… (zunubi 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
Tattaunawa
An sani cewa :
Axis na juyawa ko juyawa yana nan a wurin D.
F 1 = 10 N da l 1 = r 1 sin θ = (40 cm) (zunubi 53 o ) = (0,4 m) (0,8) = 0,32 mita
F 2 = 10√2 N da l 2 = r 2 sin θ = (20 cm) (zunubi 45 o ) = (0,2 m) (0,5√2) = 0,1√2 mita
F 3 = 20 N da l 3 = r 1 sin θ = (10 cm) (zunubi 90 o ) = (0,1 m) (1) = 0,1 mita
An tambaya : Sakamakon lokacin ƙarfi
Amsa :
τ 1 = F 1 l 1 = (10 N) (0,32 m) = 3,2 nm
(tabbatacce saboda wannan lokacin ƙarfi yana sa toshen ya juya akasin agogo)
τ 1 = F 2 l 2 = (10√2 N) (0,1√2 m) = -2 Nm
(mara kyau saboda wannan lokacin ƙarfi yana sa toshen ya juya a hannun agogo)
τ 1 = F 2 l 2 = (20 N) (0,1 m) = 2 nm
(tabbatacce saboda wannan lokacin ƙarfi yana sa toshen ya juya akasin agogo)
Lokacin ƙarfi da ya biyo baya:
Στ = τ 1 – τ 1 + τ 3
Στ = 3,2 Nm – 2 Nm + 2 Nm
Στ = 3,2 Nm
Amsar da ta dace ita ce D.
3. An sanya sandar AB, wacce aka yi watsi da nauyinta, a kwance kuma ƙarfi uku ne ke aiki da ita kamar yadda aka nuna a cikin hoton. Lokacin da ƙarfin da ke aiki a kan sandar lokacin da aka juya shi a kan axis a D shine… (zunubi 53 o = 0,8)
A. 2,4 Nm
B. 2,6 Nm
C. 3,0 Nm
D. 3,2 Nm
E. 3,4 Nm
Tattaunawa
An sani cewa :
Axis na juyawa yana nan a D.
Nisa tsakanin F 1 da kuma axis na juyawa (r AD ) = 40 cm = 0,4 m
Nisa tsakanin F 2 da kuma axis na juyawa (r BD ) = 20 cm = 0,2 m
Nisa tsakanin F 3 da kuma axis na juyawa (r CD ) = 10 cm = 0,1 m
F 1 = 10 Newton
F 2 = 10√2 Newton
F 3 = 20 Newton
Zunubi 53 o = 0,8
Tambaya : Lokacin ƙarfi da zai biyo baya idan sandar ta juya a kan axis a D
Amsa :
Lissafa lokacin ƙarfin da kowace ƙarfi ta samar.
Lokacin ƙarfi 1
Στ 1 = (F 1 )(r AD sin 53 o ) = (10 N) (0,4 m) (0,8) = 3,2 Nm
Moment of force 1 yana da kyau saboda alkiblar juyawar sandar da moment of force 1 ya haifar yana akasin agogo.
Lokacin ƙarfi 2
Στ 2 = (F 2 )(r BD zunubi 45 o ) = (10√2 N)(0,2 m)(0,5√2) = -2 Nm
Lokacin ƙarfin 2 ba shi da kyau saboda alkiblar juyawar sandar da lokacin ƙarfin 2 ya haifar tana cikin alkibla ɗaya da juyawar hannun agogo.
Lokacin ƙarfi 3
Στ 3 = (F 3 )(r CD sin 90 o ) = (20 N) (0,1 m) (1) = 2 Nm
Moment of force 3 yana da kyau saboda alkiblar juyawar sandar da moment of force 3 ya haifar yana akasin agogo.
Lokacin ƙarfi da ya biyo baya
Στ = 1 + Στ 2 + Στ 3
Στ = 3,2 – 2 + 2
Στ = mita 3,2 na Newton
Amsar da ta dace ita ce D.
Lokacin Inertia
4. Yi la'akari da hoton ƙwallo biyu da waya ta haɗa. Tsawon wayar = 12 m, l 1 = 4 m kuma an yi watsi da nauyin wayar, don haka girman lokacin inertia na tsarin shine…
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
Tattaunawa
An sani cewa :
Nauyin ƙwallon A (m A ) = 0,2 kg
Nauyin ƙwallon B (m B ) = 0,6 kg
Nisa tsakanin ƙwallon A da kuma axis na juyawa (rA ) = mita 4
Nisa tsakanin ƙwallon B da kuma axis na juyawa (r B ) = 12 – 4 = mita 8
Tambaya : Lokacin rashin ƙarfin jiki (I) na tsarin
Amsa :
Lokacin rashin kuzari na ƙwallo A
I A = (m A ) (r A 2 ) = (0,2) (4) 2 = (0,2) (16) = 3,2 kg m 2
Lokacin rashin kuzari na ƙwallo B
I B = (m B )(r B 2 ) = (0,6)(8) 2 = (0,6)(64) = 38,4 kg m 2
Lokacin inertia na tsarin barbashi :
I = I A + I B = 3,2 + 38,4 = 41,6 kg m 2
Amsar da ta dace ita ce B.
Dokar Newton ta Biyu ta Motsin Juyawa
5. Duba hoton wata takalmi mai kama da juna a gefe. Ana naɗe igiya a gefen tayoyin sannan a ja ƙarshen igiyar da ƙarfin F na 6 N. Idan nauyin tayoyin ya kai kilogiram 5 kuma radius ɗinsa ya kai cm 20, hanzarin kusurwa na tayoyin zai kasance…
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
Tattaunawa
An san cewa:
Ƙarfin taurin kai (F) = 6 Newton
Nauyin tayoyin (M) = 5 kg
Radius na tayoyi (R) = 20 cm = 20/100 m = 0,2 m
Tambaya: Hanzarin kusurwa na tayoyin (α)
Amsa:
Lissafa lokacin ƙarfi:
τ = FR = (Newton 6)(mita 0,2) = mita 1,2 na Newton
Lissafa lokacin inertia:
Tsarin da ake amfani da shi wajen auna lokacin da wani tayoyin da ke da ƙarfi a cikin siffar faifan faifai ko faranti ke aiki shine 1/2 MR 2 = 1/2 (kilogiram 5) (0,2 m) 2 = 1/2 (kilogiram 5) (0,04 m 2 ) = 1/2 (0,2) = 0,1 kg m 2.
Lissafin hanzarin kusurwa ta amfani da dabarar yanayin juyawa:
τ = ina
α = τ / I = 1,2 / 0,1 = 12 rad s -2
Amsar da ta dace ita ce E.
6. An naɗe wani abu mai ƙarfi na diski mai nauyin kilogiram 8 da radius na santimita 10 a gefen igiya tare da ɗaura nauyin kilogiram 4 a gefe ɗaya (g = 10 ms -2 ). Saurin motsi na ƙasa na kayan shine...
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
Tattaunawa
An san cewa:
Nauyin mazubin diski mai ƙarfi (m) = 8 kg
Radius na pulley mai ƙarfi (r) = 10 cm = mita 0,1
Nauyin kaya (m) = 4 kg
Saurin gudu saboda nauyi (g) = 10 m/s 2
Nauyin kaya (w) = mg = (4 kg)(10 m/s 2 ) = 40 kg m/s 2 = 40 Newtons
Tambaya: Haɓaka motsi na ƙasa na kayan aiki
Amsa:
Lissafa lokacin inertia na wani faifai mai ƙarfi:
I = 1/2 MR 2 = 1/2 (8 kg)(0,1 m) 2 = (4 kg)(0,01 m 2 ) = 0,04 kg m 2
Lissafa lokacin ƙarfi:
τ = F r = (40 N) (0,1 m) = 4 nm
Lissafin hanzarin kusurwa ta amfani da dabarar doka ta biyu ta Newton don motsi na juyawa:
Στ = I α
4 = 0,04 α
α = 4 / 0,04 = 100
Lissafa hanzarin motsi na ƙasa na kayan:
a = r α = (0,1) (100) = 10 m/s 2
Amsar da ta dace ita ce C.
7. Pulley mai ƙarfi mai nauyin (M) da radius (R) kamar yadda aka nuna a hoton! An naɗe ƙarshen igiya mara nauyi ɗaya a kusa da pulley, ɗayan ƙarshen igiyar an rataye shi da nauyin m kg, haɓaka kusurwar pulley (α) idan an saki nauyin. Idan an haɗa wani yanki na filastik A mai nauyin 1⁄2 M zuwa pulley, don samar da saurin kusurwa iri ɗaya dole ne a yi nauyin…. (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
Tattaunawa
An sani cewa :
nauyin kaya = m
Nauyin kaya = w = mg
Nauyin kura mai ƙarfi = M
Radius na pulley mai ƙarfi = R
Hanzarin kusurwa na pulley = α
An tambaya :
Idan nauyin kura ya ƙaru zuwa M + M/2 = 3M/2 kuma saurin kusurwoyin kura = α, menene nauyin kayan?
Amsa :
Lokacin inertia na kura ba tare da plasticine ba:
I = 1/2 MR 2 = 0,5 MR 2
Lokacin inertia na pulley + plasticine:
I = 1/2 (3M/2) R 2 = (3M/4) R 2 = 0,75M R 2
Lokacin ƙarfi:
τ = FR
Dokar Newton ta biyu ta motsi na juyawa:
Στ = I α
w R = I α
mg R = I α
α = mg R / I
Domin samar da irin wannan hanzarin kusurwa, dole ne a yi nauyin nauyin….. Madadin α a cikin lissafi na 2 tare da α a cikin lissafi na 1:
Amsar da ta dace ita ce B.
8.. An nuna wani kura da aka yi da wani abu mai ƙarfi wanda aka naɗe igiya a gefen waje kamar yadda aka nuna a cikin hoton. An yi watsi da gogayya ta kura da igiyar da gogayya a kan axis na juyawarta. Idan nauyin ya sauka tare da hanzari akai-akai a ms -2 , to ƙimar lokacin inertia na kura daidai yake da….
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
Tattaunawa
An san cewa:
Ƙarfi = w = mg
Hannun ƙarfi = R
Hanzarin kusurwa = α
Haɓaka kaya = a ms -2
Tambaya: Lokacin inertia na pulley (I)
Amsa:
Alaƙar da ke tsakanin hanzarin layi da hanzarin kusurwa:
a = Rα
α = a / R
Ana ƙididdige lokacin inertia ta amfani da dabarar:
τ = ina
I = τ: α = τ: a / R = τ (R / a) = τ R a -1
Babu amsar da ta dace.
9. An nuna wani kura da aka yi da wani abu mai ƙarfi wanda aka naɗe igiya a gefen waje kamar yadda aka nuna a cikin hoton. An yi watsi da gogayya ta kura. Idan lokacin inertia na kura I = β da igiyar an ja ta da ƙarfi mai ɗorewa F, to ƙimar F daidai take da….
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
Tattaunawa
An san cewa:
Ja ƙarfin = F
Lokacin inertia na pulley = β
Hanzarin kusurwa na pulley = α
Radius na kura = R
Tambaya: Darajar F daidai take da….
Amsa:
Tsarin doka ta biyu ta Newton don motsi na juyawa:
Στ = β α ———- Equation 1
Bayanin dabara:
Στ = Sakamakon lokacin ƙarfi (ƙarfi)
β = Lokacin rashin ƙarfi
α = Haɓaka kusurwa
Sakamakon lokacin ƙarfi da ke aiki akan kushin:
Στ = FR ———-> Lissafi 2
Bayanin dabara:
F = ƙarfin juriya
R = Nisa daga wurin aikin ƙarfi F zuwa ga axis na juyawa = radius na pulley
Sauya Στ a cikin equation 1 da Στ a cikin lissafi 2:
Στ = β. α
F. R = β. α
F = (β.α) / R
F = β.α. (R -1 )
Amsar da ta dace ita ce D.
Momentum na Kungular
10. Ƙwayar da ke da nauyin gram 0,2 tana motsawa a cikin da'ira tare da saurin kusurwa mai 10 rad s -1 . Idan radius na hanyar ƙwayar ta kasance 3 cm, to, saurin kusurwar ƙwayar ta kasance...
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
Tattaunawa
An san cewa:
Nauyin ƙwayoyin cuta (m) = gram 0,2 = 2 x 10 -4 kg
Gudun kusurwa (ω) = 10 rad s -1
Radius na hanyar barbashi (r) = 3 cm = mita 3 x 10 -2
An tambaya: Motsin kusurwa na barbashi
Amsa:
Tsarin ƙarfin kusurwa:
L = I ω
Bayani: I = ƙarfin kusurwa, I = lokacin inertia, ω = saurin kusurwa
Lokacin inertia na barbashi:
I = mr 2 = (2 x 10 -4 )(3 x 10 -2 ) 2 = (2 x 10 -4 )(9 x 10 -4 ) = 18 x 10 -8
Motsin kusurwa shine:
L = I ω = (18 x 10 -8 )(10 rad s -1 ) = 18 x 10 -7 kg m 2 s -1
Babu amsar da ta dace.
11. Wata mai rawa tana juyawa da hannayenta da suka miƙe har zuwa tsawon santimita 160. Sannan, ana naɗe hannayenta zuwa tsawon santimita 80 a gwiwar hannu. Idan saurin kusurwar mai rawa ya kasance iri ɗaya, to ƙarfinta na layi zai kasance...
A. har yanzu
B. ya zama 1/2 girman asali
C. ya zama 3/4 na girman asali.
D. ya zama sau biyu na asali
E. ya zama sau 4 na asali
Tattaunawa
An san cewa:
Radius 1 (r 1 ) = 160 cm
Radius 2 (r 2 ) = 80 cm
Gudun kusurwa 1 (ω 1 ) = ω
Gudun kusurwa 1 (ω 2 ) = ω
An tambaya: Momentum mai layi
Amsa:
Saurin layi 1:
v 1 = r 1 ω 1 = (160 cm) ω
Saurin layi 2:
v 2 = r 2 ω 2 = (80 cm) ω
Mitar layi ta 1:
p = mv 1 = m (160 cm) ω
Mitar layi ta 2:
p = mv 2 = m (80 cm) ω
Don haka saurin layin ya zama sau 1/2 na asali.
Amsar da ta dace ita ce B.
Tushen tambaya:
Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a
