Motsin juyawa - matsaloli da mafita

Motsin juyawa - matsaloli da mafita

Torque

1. Tsawon katako mai tsawon santimita 140. Akwai ƙarfi uku a kan katakon, F ₁ = 20 N, F ₂ = 10 N, da F ₃ = 40 N tare da alkibla da matsayi kamar yadda aka nuna a cikin hoton da ke ƙasa. Menene ƙarfin juyi da ke sa katakon ya juya a tsakiyar nauyin katakon?

An sani:

Cibiyar taro tana tsakiyar katakon.

Tsawon katako (l) = 140 cm = mita 1.4

Ƙarfi 1 (F ₁ ) = 20 N, hannun lever 1 (l ₁ ) = 70 cm = mita 0.7

Ƙarfi 2 (F ₂ ) = 10 N, hannun lever 2 (l ₂ ) = 100 cm – 70 cm = 30 cm = mita 0.3

Ƙarfi 3 (F ₃ ) = 40 N, hannun lever 3 (l ₃ ) = 70 cm = mita 0.7

Ana so: Girman karfin juyi

Magani:

Juyin juyi na 1 yana juya hasken a hannun agogo, don haka an sanya alamar korau ga juyin juyi na 1.

τ ₁ = F ₁ l ₁ = (20 N)(0.7 m) = -14 N m

Juyin juyi na 2 yana juya katakon a akasin agogon, don haka yana sanya alama mai kyau ga juyin juyi na 2.

τ ₂ = F ₂ l ₂ = (10 N)(0.3 m) = 3 N m

Juyin juya 3 yana juyawa ta hannun agogo, don haka an sanya alama mai kyau ga juyin juya 3.

τ ₃ = F ₃ l3 = (40 N)(0.7 m) = -28 N m

Ƙarfin wutar lantarki:

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

Girman karfin juyi shine 39 N m. Alkiblar juyawar hasken wutar lantarki ta hannun agogo, don haka an sanya alama mara kyau.

2. Menene ƙarfin wutar lantarki mai ƙarfi ke aiki akan hasken. Axis na juyawa a wurin D. (sin 53 ^o = 0.8)

An sani:

Axis na juyawa 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 zunubi 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

Ana so: Ƙarfin wutar lantarki

Magani:

τ ₁ = F ₁ l ₁ = (10 N) (0.32 m) = 3.2 nm

(Juyin juya 1 yana juya katako a akasin agogo don haka muna sanya alama mai kyau ga juyin juya 1)

τ ₂ = F ₂ l ₂ = (10√2 N) (0.1√2 m) = -2 Nm

(Juyin juya 2 yana juya haske a hannun agogo don haka muna sanya alamar korau ga juyin juya 2)

τ ₃ = F ₂ l ₂ = (20 N) (0.1 m) = 2 nm

(Juyin juya 3 yana juya katako a akasin agogo don haka muna sanya alama mai kyau ga juyin juya 3)

Ƙarfin wutar lantarki:

Στ = τ ₁ – τ ₁ + τ ₃

Στ = 3.2 Nm – 2 Nm + 2 Nm

Στ = 3.2 Nm

3. Menene karfin juyi na net idan axis na juyawa a wurin D. (sin 53 ^o = 0.8)

An sani:

Axis na juyawa a wurin 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

Ana so: Ƙarfin wutar lantarki

Magani:

Lokacin ƙarfin 1

Στ ₁ = (F ₁ )(r _AD sin 53 ^o ) = (10 N) (0.4 m) (0.8) = 3.2 Nm

(Juyin juya 1 yana juya katako a akasin agogo don haka muna sanya alama mai kyau ga juyin juya 1)

Lokacin ƙarfin 2

Στ ₂ = (F ₂ )(r _BD zunubi 45 ^o ) = (10√2 N)(0.2 m)(0.5√2) = -2 Nm

(Juyin juya 2 yana juya haske a hannun agogo don haka muna sanya alamar korau ga juyin juya 2)

Lokacin ƙarfin 3

Στ ₃ = (F ₃ )(r _CD sin 90 ^o ) = (20 N) (0.1 m) (1) = 2 Nm

(Juyin juya 2 yana juya katako a akasin agogo don haka muna sanya alama mai kyau ga juyin juya 3)

Ƙarfin wutar lantarki:

Στ = ₁ + Στ ₂ + Στ ₃

Στ = 3.2 – 2 + 2

Στ = mita 3.2 na Newton

Lokacin inertia

4. Tsawon waya = 12 m, l ₁ = 4 m. Yi watsi da nauyin waya. Menene lokacin rashin ƙarfin tsarin.

An sani:

Nauyin A (mA ₎ = 0.2 kg

Nauyin B (m _B ) = 0.6 kg

Nisa tsakanin A da kuma axis na juyawa (rA ₎ = mita 4

Nisa tsakanin B da kuma axis na juyawa (r _B ) = 12 – 4 = mita 8

Ana so: Lokacin rashin ƙarfin tsarin

Magani:

Lokacin inertia na A

I _A = (m _A ) (r _A ² ) = (0.2) (4) ² = (0.2) (16) = 3.2 kg m ²

Lokacin inertia na B

I _B = (m _B )(r _B ² ) = (0.6)(8) ² = (0.6)(64) = 38.4 kg m ²

Lokacin inertia na tsarin:

I = I _A + I _B = 3.2 + 38.4 = 41.6 kg m ²

Tsarin juyawa

5. Ana amfani da ƙarfin 6-N a kan igiya da aka naɗe a kusa da kura mai nauyin M = 5 kg da radius R = 20 cm. Menene hanzarin kusurwa na kura. Kura mai ƙura silinda ce mai ƙarfi iri ɗaya.

An sani:

Ƙarfi (F) = 6 Newton

Nauyi (M) = 5 kg

Radius (R) = 20 cm = 20/100 m = 0.2 m

Ana so: Hanzarin kusurwa (α)

Magani:

Lokacin ƙarfin:

τ = FR = (6 Newton) (0.2 mita) = 1.2 N m

Lokacin inertia ga silinda mai ƙarfi:

I = 1/2 MR ²

I = 1/2 (kilogiram 5)(0.2 m) ²

I = 1/2 (kilogiram 5)(0.04 m ² )

I = 1/2 (0.2)

I = 0.1 kg ^m2.

Haɓakar kusurwa:

τ = ina

α = τ / I = 1.2 / 0.1 = 12 rad s ^-2

6. Toshe mai nauyi = kilogiram 4 da aka rataye daga igiya da aka naɗe a kusa da kura mai nauyi = kilogiram 8 da radius R = 10 cm. Saurin da aka samu sakamakon nauyi shine 10 ms ^-2 . Menene saurin da aka samu a layi na toshe? Kura mai nauyi silinda ce mai ƙarfi iri ɗaya.

An sani:

Nauyin kura (m) = 8 kg

Radius na kura (r) = 10 cm = 0.1 m

Nauyin tubali (m) = 4 kg

Saurin gudu saboda nauyi (g) = 10 m/s ²

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

Ana so: Haɓaka faɗuwar tubalan kyauta

Magani:

Lokacin inertia na silinda 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

Lokacin ƙarfin:

τ = F r = (40 N) (0.1 m) = 4 nm

Haɓakar kusurwa:

Στ = I α

4 = 0.04 α

α = 4 / 0.04 = 100

Haɓaka layi:

a = r α = (0.1) (100) = 10 m/s ²

7. Toshe mai nauyin m da aka rataye daga igiya da aka naɗe a kusa da pulley. Idan saurin faɗuwar toshewar a hankali shine am/s ² , menene lokacin inertia na pulley..

An sani:

nauyi = w = mg

Hannun Lever = R

Haɓakar kusurwa = α

Haɓaka faɗuwar free fall na toshe = a ms ^-2

Ana so: Lokacin inertia na pulley (I)

Magani:

Alaƙa tsakanin hanzarin layi da hanzarin kusurwa:

a = Rα

α = a / R

Lokacin inertia:

τ = ina

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

Motsin kusurwa

8. Kwayar gram 0.2 tana motsawa a cikin da'ira a gudun da ya dace na m10/s. Radius na da'irar shine cm 3. Menene ma'aunin kusurwa na ƙwayar?

An sani:

Nauyin barbashi (m) = gram 0.2 = 2 x 10 ^-4 kg

Gudun kusurwa (ω) = 10 rad s ^-1

Radius (r) = 3 cm = mita 3 x 10 ^-2

Ana so: Motsin kusurwa na ƙwayar

Magani:

Daidaiton ƙarfin kusurwa:

L = I ω

I = ƙarfin kusurwa, I = lokacin inertia, ω = saurin kusurwa

Lokacin inertia (ga ƙwayoyin cuta):

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:

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

1. Menene motsin juyawa?
   • Amsa: Motsin juyawa yana nufin motsin abu a kusa da wani tsayayyen axis. Wannan nau'in motsi ne wanda kowane wuri na abu ke motsawa a cikin da'ira a kusa da axis.
2. Ta yaya saurin layi yake da alaƙa da saurin kusurwa a cikin motsi na juyawa?
   • Amsa: Saurin layi ($v$) na wani wuri a cikin abin da ke juyawa yana daidai gwargwado kai tsaye da nisansa ($r$) daga axis na juyawa da kuma saurin kusurwa ($\omega$) na abu. An bayar da dangantakar ta $v=r\cdot \omega$.
3. Menene lokacin inertia, kuma ta yaya yake da alaƙa da motsi na juyawa?
   • Amsa: Moment of inertia shine kwatancen juyawa na taro a cikin motsi na layi. Yana auna juriyar abu ga canje-canje a yanayin juyawarsa. Lokacin inertia ya dogara da nauyin abu da kuma rarrabawarsa dangane da axis na juyawa.
4. Ta yaya dokar farko ta motsi ta Newton ta shafi motsin juyawa?
   • Amsa: Kamar yadda abu a cikin motsi na layi yake ci gaba da motsi sai dai idan wani ƙarfi na waje ya yi masa aiki, abu a cikin motsi na juyawa zai ci gaba da kasancewa a cikin wannan yanayin sai dai idan ƙarfin juyi na waje ya yi masa aiki.
5. Menene mahimmancin radius na gyration?
   • Amsa: Radius na gyration yana ba da ma'aunin rarraba nauyin abu daga axis na juyawarsa. Ainihin yana bayyana yadda dukkan nauyin abu zai buƙaci a taru daga axis don samun lokaci ɗaya na inertia kamar rarrabawar asali.
6. Menene ƙarfin angular kuma ta yaya ake kiyaye shi?
   • Amsa: Motsin kusurwa daidai yake da juyawar motsi na layi. Sakamakon lokacin inertia na abu ne da saurin kusurwarsa. A cikin tsarin rufewa, jimlar motsin kusurwa yana ci gaba da kasancewa ba tare da an yi aiki da shi ba sai dai idan ƙarfin juyi na waje ya yi aiki da shi, wanda ke nuna kiyayewar motsin kusurwa.
7. Ta yaya karfin juyi ke tasiri ga motsin juyawa?
   • Amsa: Juyawa daidai yake da ƙarfin juyawa. Yana haifar da canje-canje a cikin motsin juyawa na abu. An bayar da alaƙar ta hanyar dokar Newton ta biyu don juyawa: $\tau =I\cdot \alpha$, inda $\tau$ shine ƙarfin juyi, $I$ shine lokacin inertia, kuma $\alpha$ shine hanzarin kusurwa.
8. Ta yaya tsakiyar taro ya bambanta da tsakiyar juyawa?
   • Amsa: Duk da cewa suna iya haɗuwa, tsakiyar taro shine wurin da za a iya ɗauka cewa dukkan nauyin abu ya taru don dalilai na lissafi a cikin motsi na layi, yayin da tsakiyar juyawa shine wurin (ko axis) wanda abu ke juyawa akai.
9. Menene rawar da ƙarfin centripetal ke takawa a cikin motsin juyawa?
   • Amsa: Ƙarfin tsakiya shine ƙarfin da ke aiki akan abu da ke tafiya a cikin hanyar da'ira, wanda aka nuna zuwa tsakiyar juyawa. Yana da alhakin kiyaye abu a cikin hanyarsa mai lanƙwasa da hana shi motsi a cikin layi madaidaiciya saboda rashin ƙarfi.

10. Ta yaya makamashin motsi na juyawa yake da alaƙa da lokacin inertia da saurin kusurwa?

    • Amsa: Ƙarfin motsi na juyawa shine kuzarin da ke faruwa sakamakon juyawar abu a kusa da wani axis. An bayar da shi ta hanyar dabarar: KE_{rotational​}=1/2 ​I ω², inda $I$ shine lokacin inertia da kuma $\omega$ shine saurin kusurwa.