Misali na lokacin inertia

Misalan Tambayoyi 25 na Lokacin Rashin Inertia

Lokacin Rashin Inertia na Barbashi

1. An haɗa ƙwallon da nauyinta ya kai gram 100 da igiya mai tsawon santimita 30 kamar yadda aka nuna a hoton. Lokacin inertia ƙwallon da ke kewaye da axis AB shine...
Tattaunawa
An san cewa:
Axis na juyawa shine AB
Nauyin ƙwallon (m) = gram 100 = 100/1000 = kilogiram 0,1
Nisa daga ƙwallo daga ma'aunin juyawa (r) = 30 cm = mita 0,3
An tambaya: Lokacin rashin kuzari na ƙwallon (I)
Amsa:
Ni = Mr.² = (0,1 kg)(0,3 m)²
I = (0,1 kg)(0,09 m²)
I = 0,009 kg m²

2. Nauyin ƙwallon m₁ shine gram 100 kuma nauyin ƙwallon shine m₂ Nauyin ya kai gram 200. An haɗa ƙwallan biyu da waya mai tsawon santimita 60 tare da nauyin da ba shi da yawa. Axis ɗin AB yana tsakiyar wayar. Lokacin inertia na ƙwallan biyu game da axis ɗin AB shine…
Tattaunawa
An san cewa:
Nauyin ƙwallon 1 (m)₁) = gram 100 = 100/1000 = kilogiram 0,1
Nisa na ball 1 daga axis na juyawa (r)₁) = 30 cm = 30/100 = mita 0,3
Nauyin ƙwallon 2 (m)₂) = gram 200 = 200/1000 = kilogiram 0,2
Nisa na ball 2 daga axis na juyawa (r)₂) = 30 cm = 30/100 = mita 0,3
An tambaya: Lokacin rashin ƙarfi na tsarin ƙwallo biyu
Amsa:
Ni = m₁ r₁² +m₂ r₂²
I = (0,1 kg)(0,3 m)² + (0,2 kg)(0,3 m)²
I = (0,1 kg)(0,09 m²) + (0,2 kg)(0,09 m²)
I = 0,009 kg m² + 0,018 kg m²
I = 0,027 kg m²

3. Nauyin ƙwallon m₁ shine gram 200 kuma nauyin ƙwallon shine m₂ gram 100 ne. An haɗa ƙwallan biyu ta hanyar waya mai tsawon santimita 60 kuma ba a kula da nauyinta ba. Axis na AB yana kan ƙwallon m.₂Lokacin inertia na tsarin ƙwallo biyu game da axis na AB shine…
Tattaunawa
An san cewa:
Nauyin ƙwallon 1 (m)₁) = gram 200 = 200/1000 = kilogiram 0,2
Nisa na ball 1 daga axis na juyawa (r)₁) = 60 cm = 60/100 = mita 0,6
Nauyin ƙwallon 2 (m)₂) = gram 100 = 100/1000 = kilogiram 0,1
Nisa na ball 2 daga axis na juyawa (r)₂) = mita 0
An tambaya: Lokacin rashin ƙarfi na tsarin ƙwallo biyu
Amsa:
Ni = m₁ r₁² +m₂ r₂²
I = (0,2 kg)(0,6 m)² + (0,2 kg)(0)²
I = (0,2 kg)(0,36 m²) + 0
I = 0,072 kg m²

4. Kowane ƙwallon yana da nauyin gram 100 kuma an haɗa shi da waya mai nauyin da ba a saba gani ba. Tsawon wayar shine 60 cm kuma faɗin shine 30 cm. A ƙayyade lokacin inertia na tsarin ƙwallon game da axis na AB…
Tattaunawa
An san cewa:
Nauyin ƙwallon 1 (m)₁) = m₂ = m₃ = m₄ = gram 100 = 100/1000 = kilogiram 0,1
Nisa na ball 1 daga axis na juyawa (r)₁) = 30 cm = 30/100 = mita 0,3
Nisa na ball 2 daga axis na juyawa (r)₂) = 30 cm = 30/100 = mita 0,3
Nisa na ball 3 daga axis na juyawa (r)₃) = 30 cm = 30/100 = mita 0,3
Nisa na ball 4 daga axis na juyawa (r)₄) = 30 cm = 30/100 = mita 0,3
An tambaya: Lokacin rashin ƙarfin tsarin ƙwallon
Amsa:
Ni = m₁ r₁² +m₂ r₂² +m₃ r₃² +m₄ r₄²
I = (0,1 kg)(0,3 m)² + (0,1 kg)(0,3 m)² + (0,1 kg)(0,3 m)² + (0,1 kg)(0,3 m)²
I = (0,1 kg)(0,09 m²) + (0,1 kg)(0,09 m²) + (0,1 kg)(0,09 m²) + (0,1 kg)(0,09 m²)
I = 0,036 kg m²

5. Duba hoton da ke ƙasa. Ƙwayoyin cuta guda huɗu, kowannensu yana da nauyin mita 4 (a nisan r daga tsakiya), mita 3 (a nisan r daga tsakiya), mita 2 (a nisan 2r daga tsakiya), mita 2 (a nisan r daga tsakiya). Tsarin yana cikin jirgin xy. Idan tsarin yana juyawa a kusa da axis na x, to lokacin inertia na tsarin shine…
A. 5 m r
B. 7 m r     
C. 5 m r²
D. 6 m r²
E. 7 m r²

Tattaunawa:
An sani :
An san cewa:
m₁ = mita 4 (nisa daga tsakiya), m₂ = mita 3 (nisa daga tsakiya), m₃ = 2m (nisa 2r daga tsakiya), m₄ = 2m (nisa daga tsakiya)
An tambaya :
Lokacin rashin kuzari (I)?
Jawab :
Tsarin lokacin inertia na barbashi :
Ni = Mr.²
Ana juya tsarin a kusa da axis na x don haka m₂ kuma m₄ yana kan axis na juyawa. Domin yana kan axis na juyawa, r₂ kuma r₄ darajar sifili ce.
I₁ = m₁ r₁² = 4m²
I₂ = m₂ r₂² = mita 3 (0)² = 0
I₃ = m₃ r₃² = mita 2 (2r)² = mita 2 (4r)²) = 8mr²
I₄ = m₄ r₄² = mita 2 (0)² = 0
Lokacin rashin ƙarfin tsarin :
Ni = Ni₁ + Ni₂ + Ni₃ + Ni₄
I = 4mr² + 8m²
I = 12mr²

6. Rod AB yana da nauyin kilogiram 3 idan aka juya shi ta cikin B, lokacin inertia ɗinsa shine kilogiram 27 m²Idan aka juya ta cikin C, lokacin inertia zai zama…
A. 70 kg m²
B. 76 kg m²
C. 92 kg m²
D. 98 kg m²
E. 108 kg m²
Tattaunawa:

An sani :
m = 3 kg
I_B = 27 kg m²
An tambaya :
I_C ?
Jawab :
I_B = Mr.²
27 kg m² = (kilogiram 3)(r²)
27 kg m² / 3 kg = r²
9 m² =r²
r = 3 m
Tsawon AB = tsawon AC = mita 3

Lokacin Inertia sandar idan aka juya ta cikin C:
I_C = Mr.²
I_C = (kilogiram 3)(mita 6)²
I_C = (kilogiram 3)(36 m²) = 108 kg m²

7.

An haɗa ƙwallaye biyu, waɗanda ake ɗauka a matsayin ƙwayoyin cuta, ta hanyar igiyar waya kamar yadda aka nuna a cikin hoton. Idan yawan ƙwallayen P da Q sun kai gram 600 da gram 400 bi da bi, to lokacin inertia na tsarin ƙwallayen biyu game da axis AB shine…

A. 0,008 kg.m²
B. 0,076 kg.m²
C. 0,124 kg.m²
D. 0,170 kg.m²
E. 0,760 kg.m²

Tattaunawa
An sani :
Nauyin ƙwallon P (m)_P) = gram 600 = 0,6 kg
Nauyin ƙwallon Q (m)_Q) = gram 400 = 0,4 kg
Nisa na ƙwallon P daga axis na juyawa (r)_P) = 20 cm = mita 0,2
Nisa na ƙwallon Q daga axis na juyawa (r)_Q) = 50 cm = mita 0,5
An tambaya : Lokacin inertia (I) tsarin dangane da axis ko axis na juyawa AB
Jawab :

Tsarin lokacin inertia na barbashi :
Ni = Mr.²
Bayani: I = lokacin inertia, m = yawan barbashi, r = nisan barbashi daga axis na juyawa
Lokacin rashin kuzari na ƙwallon P
I_P = (m_P)(r_P²) = (0,6)(0,2)² = (0,6)(0,04) = 0,024 kg m²
Lokacin rashin kuzari na ƙwallon Q
I_Q = (m_Q)(r_Q²) = (0,4)(0,5)² = (0,4)(0,25) = 0,1 kg m²
Lokacin inertia na tsarin barbashi :
Ni = Ni_P + Ni_Q = 0,024 + 0,1 = 0,124 kg m²
Amsar da ta dace ita ce C.

8.

An shirya ƙwallaye biyu da waya ta haɗa (an yi watsi da nauyin wayar) kamar yadda aka nuna a cikin hoton. Girman lokacin inertia ɗinsu shine…

A. 20 x 10^-3 kg.m²
B. 25 x 10^-3 kg.m²
C. 11 x 10^-2 kg.m²
D. 55 x 10^-2 kg.m²
E. 80 x 10^-2 kg.m²

Tattaunawa
An sani :
Nauyin ƙwallon A (m)_A) = gram 200 = 0,2 kg
Nauyin ƙwallon B (m)_B) = gram 400 = 0,4 kg
Nisa tsakanin ƙwallon A da kuma axis na juyawa (r)_A) = 0
Nisa tsakanin ƙwallon B da kuma axis na juyawa (r)_B) = 25 cm = mita 0,25
An tambaya Lokacin rashin kuzari (I) na tsarin
Jawab :
Lokacin rashin kuzari na ƙwallo A
I_A = (m_A)(r_A²) = (0,2)(0)² = 0
Lokacin rashin kuzari na ƙwallo B
I_B = (m_B)(r_B²) = (0,4)(0,25)² = (0,4)(0,0625) = 0,025 kg m²
Lokacin inertia na tsarin barbashi :
Ni = Ni_A + Ni_B = 0 + 0,025 = 0,025 kg m² = 25 x10^-3 kg m²
Amsar da ta dace ita ce B.

9.
Kalli hoton ƙwallo biyu da aka haɗa ta waya. Tsawon waya = 12 m, l₁ = 4 m kuma an yi watsi da nauyin wayar, to girman lokacin inertia na tsarin shine…
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
Tattaunawa
An sani :
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 (r)_A) = mita 4
Nisa tsakanin ƙwallon B da kuma axis na juyawa (r)_B) = 12 – 4 = mita 8
An tambaya Lokacin rashin kuzari (I) na tsarin
Jawab :
Lokacin rashin kuzari na ƙwallo A
I_A = (m_A)(r_A²) = (0,2)(4)² = (0,2)(16) = 3,2 kg m²
Lokacin rashin kuzari na ƙwallo B
I_B = (m_B)(r_B²) = (0,6)(8)² = (0,6)(64) = 38,4 kg m²
Lokacin inertia na tsarin barbashi :
Ni = Ni_A + Ni_B = 3,2 + 38,4 = 41,6 kg m²
Amsar da ta dace ita ce B.

Lokacin Rashin Inertia na Jiki Mai Tauri

10. Sanda mai ƙarfi tana da nauyin kilogiram 2 da tsawon mita 2. Ka ƙayyade lokacin inertia na sandar idan axis na juyawa yana tsakiyar sandar!
Tattaunawa
An san cewa:
Nauyin sandar ƙarfi (M) = 2 kg
Tsawon tushe mai ƙarfi (L) = mita 2
An tambaya: Lokacin inertia
Amsa:
Tsarin da ake amfani da shi wajen auna lokacin da sandar take juyawa idan axis na juyawa yana tsakiyar sandar:
I = (1/12) ML²
I = (1/12) (2 kg)(2 m)²
I = (1/12) (2 kg)(4 m)²)
I = (1/12)(8 kg m²)
I = 8/12 kg m²
I = 2/3 kg m²

11. Sanda mai ƙarfi tana da nauyin kilogiram 2 da tsawon mita 2. Ka ƙayyade lokacin inertia na sandar idan axis na juyawa yana a ƙarshen sandar!
Tattaunawa
An san cewa:
Nauyin sandar ƙarfi (M) = 2 kg
Tsawon tushe mai ƙarfi (L) = mita 2
An tambaya: Lokacin inertia
Amsa:
Tsarin da ake amfani da shi wajen auna lokacin da sandar take juyawa idan axis na juyawa yana a ƙarshen sandar:
I = (1/3) ML²
I = (1/3) (2 kg)(2 m)²
I = (1/3) (2 kg)(4 m)²)
I = (1/3)(8 kg m²)
I = 8/3 kg m²

12. Kayyade lokacin rashin ƙarfin diski mai ƙarfi mai nauyin kilogiram 10 da radius na mita 0,1, idan axis na juyawa yana tsakiyar diskin, kamar yadda aka nuna a cikin hoton!
Tattaunawa
An san cewa:
Nauyin faifai mai ƙarfi (M) = 10 kg
Radius ɗin diski mai ƙarfi (L) = mita 0,1
An tambaya: Lokacin rashin ƙarfi na faifan diski mai ƙarfi
Amsa:
Tsarin lokacin inertia na sandar idan axis na juyawa yana tsakiyar faifan:
I = (1/2) ML²
I = (1/2) (10 kg)(0,1 m)²
I = (1/2) (10 kg)(0,01 m)²)
I = (1/2)(0,1 kg m²)
I = 0,05 kg m²

13. Kayyade lokacin rashin ƙarfin ƙwallon da ta yi ƙarfi mai nauyin kilogiram 20 da kuma radius na mita 0,1, idan axis na juyawa yana tsakiyar ƙwallon, kamar yadda aka nuna a cikin hoton!
Tattaunawa
An san cewa:
Nauyin ƙwallon da aka yi da ƙarfi (M) = 20 kg
Radius na ƙwallon da tauri (L) = mita 0,1
An tambaya: Lokacin inertia
Amsa:
Tsarin da ake amfani da shi wajen auna lokacin da sandar take juyawa idan axis na juyawa yana tsakiyar wani yanki mai ƙarfi:
I = (2/5) ML²
I = (2/5)(20 kg)(0,1 m)²
I = (2/5)(20 kg)(0,01 m²)
I = (2/5)(0,2 kg m²)
I = 0,4/5 kg m²
I = 0,08 kg m²

14. Kayyade lokacin rashin ƙarfin ƙwallon siriri mai rami mai nauyin kilogiram 0,5 da kuma radius na mita 0,1, idan axis na juyawa yana tsakiyar ƙwallon, kamar yadda aka nuna a cikin hoton!
Tattaunawa
An san cewa:
Nauyin ƙwallon siriri (M) = 0,5 kg
Radius na siraran ƙwallo (L) = mita 0,1
An tambaya: Tsarin lokacin inertia
Amsa:
Tsarin da ake amfani da shi wajen auna lokacin da sandar take juyawa idan axis na juyawa yana tsakiyar wani siririn ƙwallo:
I = (2/3) ML²
I = (2/3) (0,5 kg)(0,1 m)²
I = (2/3) (0,5 kg)(0,01 m)²)
I = (2/3)(0,005 kg m²)
I = 0,01/3 kg m²

15. Farantin mai kauri mai kusurwa huɗu yana da nauyin kilogiram 2, tsawonsa mita 0,5 da faɗinsa mita 0,2. Ka ƙayyade lokacin inertia na farantin mai kauri idan axis na juyawa yana tsakiyar farantin, kamar yadda aka nuna a cikin hoton!
Tattaunawa
An san cewa:
Nauyin farantin murabba'i mai siffar murabba'i (M) = 2 kg
Tsawon faranti (a) = mita 0,5
Faɗin faranti (b) = mita 0,2
An tambaya: Tsarin lokacin inertia
Amsa:
Tsarin lokacin inertia na farantin mai ƙarfi mai kusurwa huɗu idan axis na juyawa yana tsakiyar farantin mai ƙarfi:
I = (1/12) M (a)² + b²)
I = (1/12)(2)(0,52 + 0,22)
I = (2/12)(0,25 + 0,04)
I = (1/6)(0,29)
I = 0,29/6 kg m²

16. Ga bayanin da ke ƙasa game da abubuwan da ke haifar da motsin juyawa.

(1) Gudun kusurwa

(2) Matsayin axis na juyawa

(3) Siffar abin

(4) Nauyin abu

Abubuwan da ke tasiri ga girman lokacin inertia sune ...

A. (1), (2), (3) da (4)

B. (1), (2) da (3)

C. (1), (3) da (4)

D. (2), (3) da (4)

E. (2) da (4) kawai

Tattaunawa

Tsarin lokacin inertia:

Ni = Σm r²

Bayani: m = nauyin abu, r = nisan daga axis na juyawa.

Bisa ga wannan dabarar, an kammala da cewa lokacin inertia yana tasiri ne ta hanyar matsayin juyawar axis (2) da nauyin abu (4).

Amsar da ta dace ita ce E.

17. An haɗa ƙwallaye biyu, waɗanda ake ɗauka a matsayin ƙwayoyin cuta, ta hanyar igiyar waya kamar yadda aka nuna a cikin hoton. Idan yawan ƙwallayen P da Q sun kai gram 600 da gram 400 bi da bi, to lokacin inertia na tsarin ƙwallayen biyu game da axis AB shine…

A. 0,008 kg m²

B. 0,076 kg m²

C. 0,124 kg m²

D. 0,170 kg m²

E. 0,760 kg m²

Tattaunawa

An sani :

Axis na juyawa shine AB.

m_p = gram 600 = 0,6 kg, m_q = gram 400 = 0,4 kg

r_p = 20 cm = 0,2 m, r_q = 50 cm = 0,5 m

An tambaya : Lokacin rashin ƙarfin tsarin?

Jawab :

Ni = m_p r_p² +m_q r_q²

I = (0,6 kg)(0,2 m)² + (0,4 kg)(0,5 m)²

I = (0,6 kg)(0,04 m²) + (0,4 kg)(0,25 m²)

I = 0,024 kg m² + 0,1 kg m²

I = 0,124 kg m²

Amsar da ta dace ita ce C.

18. Lokacin inertia na wani abu yana juyawa a kusa da wani wuri mai tsayayye yana tasiri ta hanyar...

A. nauyin abu

B. girman abin

C. yawan abu

D. hanzarta juyawa ta kusurwa

E. saurin kusurwa na farko

Tattaunawa

Tsarin lokacin inertia:

Ni = ∑mr²

Bayani: I = lokacin inertia, m = taro, r = nisa daga axis na juyawa

Amsar da ta dace ita ce A.

19. Rod AB mai nauyin kilogiram 2 ana juyawa ta cikin maki A kuma lokacin inertia ɗinsa shine kilogiram 8 m²Idan aka juya ta tsakiyar wurin O (AO = OB), lokacin inertia zai zama...

A. 2 kg m²

B. 4 kg m²

C. 8 kg m²

D. 12 kg m²

E. 16 kg m²

Tattaunawa

An san cewa:

Nauyin sanda AB (m) = 2 kg

Idan aka juya ta cikin maki A ta yadda radius na juyawa (r) = nisa AB = r to lokacin inertia (I) = 8 kg m²

An tambaya: Idan aka juya ta cikin ma'aunin O ta yadda radius na juyawa (r) = nisa AO = nisa OB = 1/2 r to lokacin inertia zai zama (I) = ……

Amsa:

Ni = Mr.²

8 kg m² = (2 kg) r²

8 m² = (2) r²

r² = 8 m² / 2

r² = 4 m²

r = mita 2

Idan aka juya ta cikin ma'aunin O ta yadda 1/2 r = mita 1, to lokacin inertia zai zama:

Ni = Mr.² = (2 kg)(1 m)² = (kilogiram 2)(1 m²) = 2 kg m²

Amsar da ta dace ita ce A.

20. An haɗa ƙwallaye biyu, waɗanda ake ɗauka a matsayin ƙwayoyin cuta, ta hanyar igiyar waya kamar yadda aka nuna a cikin hoton. Idan yawan ƙwallayen P da Q sun kai gram 600 da gram 400 bi da bi, to lokacin inertia na tsarin ƙwallayen biyu game da axis AB shine…

A. 0,008 kg.m²

B. 0,076 kg.m²

C. 0,124 kg.m²

D. 0,170 kg.m²

E. 0,760 kg.m²

Tattaunawa

An sani :

Nauyin ƙwallon P (m)_P) = gram 600 = 0,6 kg

Nauyin ƙwallon Q (m)_Q) = gram 400 = 0,4 kg

Nisa na ƙwallon P daga axis na juyawa (r)_P) = 20 cm = mita 0,2

Nisa na ƙwallon Q daga axis na juyawa (r)_Q) = 50 cm = mita 0,5

An tambaya : Lokacin inertia (I) na tsarin game da axis ko juyawar axis AB

Jawab :
Tsarin lokacin inertia na barbashi :

Ni = Mr.²

Bayani: I = lokacin inertia, m = yawan barbashi, r = nisan barbashi daga axis na juyawa

Lokacin rashin kuzari na ƙwallon P

I_P = (m_P)(r_P²) = (0,6)(0,2)² = (0,6)(0,04) = 0,024 kg m²

Lokacin rashin kuzari na ƙwallon Q

I_Q = (m_Q)(r_Q²) = (0,4)(0,5)² = (0,4)(0,25) = 0,1 kg m²

Lokacin inertia na tsarin barbashi :

Ni = Ni_P + Ni_Q = 0,024 + 0,1 = 0,124 kg m²

Amsar da ta dace ita ce C.

21. An shirya ƙwallaye biyu da waya ta haɗa (an yi watsi da nauyin wayar) kamar yadda aka nuna a cikin hoton. Girman lokacin inertia ɗinsu shine…

A. 20 x 10^-3 kg.m²

B. 25 x 10^-3 kg.m²

C. 11 x 10^-2 kg.m²

D. 55 x 10^-2 kg.m²

E. 80 x 10^-2 kg.m²

Tattaunawa

An sani :

Nauyin ƙwallon A (m)_A) = gram 200 = 0,2 kg

Nauyin ƙwallon B (m)_B) = gram 400 = 0,4 kg

Nisa tsakanin ƙwallon A da kuma axis na juyawa (r)_A) = 0

Nisa tsakanin ƙwallon B da kuma axis na juyawa (r)_B) = 25 cm = mita 0,25

An tambaya Lokacin rashin kuzari (I) na tsarin

Jawab :

Lokacin rashin kuzari na ƙwallo A

I_A = (m_A)(r_A²) = (0,2)(0)² = 0

Lokacin rashin kuzari na ƙwallo B

I_B = (m_B)(r_B²) = (0,4)(0,25)² = (0,4)(0,0625) = 0,025 kg m²

Lokacin inertia na tsarin barbashi :

Ni = Ni_A + Ni_B = 0 + 0,025 = 0,025 kg m² = 25 x10^-3 kg m²

Amsar da ta dace ita ce B.

22. Barbashi huɗu na taro daban-daban suna kan layi ɗaya, kamar yadda aka nuna a cikin hoton. Lokacin inertia na tsarin tare da axis na kwance p shine…

A. 17 mb²

B. 22 mb²

C. 27 mb²

D. 31 mb²

E. 33 mb²

Tattaunawa

Axis na juyawa = layin kwance p

An san cewa:

Nauyin barbashi A (m)_A) = m

Nauyin barbashi B (m)_B) = mita 2

Nauyin ƙwayar cuta C (m)_C) = mita 3

Nauyin barbashi D (m)_D) = mita 4

Nisa tsakanin barbashi A da juyawar (r)_A) = b

Nisa tsakanin barbashi B da axis na juyawa (r)_B) = b

Nisa tsakanin barbashi C da axis na juyawa (r)_C) = 2b

Nisa tsakanin barbashi D daga axis na juyawa (r)_D) = 2b

An tambaya: Lokacin inertia na tsarin tare da axis na kwance p

Amsa:

Ni = m_A r_A² +m_B r_B² +m_C r_C² +m_D r_D²

I = (m)(b)² + (2m)(b)² + (3m)(2b)² + (4m)(2b)²

I = mb² + 2 mb² + (3m)(4b)²) + (4m)(4b)²)

I = mb² + 2 mb² + 12 mb² + 16 mb²

I = 31 mb²

Amsar da ta dace ita ce D.

23. Kalli hoton da ke ƙasa!

An haɗa ƙwayoyin cuta guda huɗu da sandar da ba ta da wani nauyi. Lokacin inertia na tsarin da ke kewaye da axis yana ta hanyar m₁ kuma m₂ shine…

A. 1/200 kg.m²

B. 1/300 kg.m²

C. 1/400 kg.m²

D. 1/600 kg.m²

E. 1/800 kg.m²

Tattaunawa

An san cewa:

Nauyin barbashi 1 (m₁) = 1/4 kg

Nauyin barbashi 2 (m₂) = 1/2 kg

Nauyin barbashi 3 (m₃) = 1/4 kg

Nauyin barbashi 4 (m₄) = 1/4 kg

Nisa tsakanin barbashi 1 da axis na juyawa (r)₁) = 0

Nisa tsakanin barbashi 2 da axis na juyawa (r)₂) = 0

Nisa tsakanin barbashi 3 da axis na juyawa (r)₃) = 10 cm = 10/100 mita = 1/10 mita

Nisa tsakanin barbashi 4 da axis na juyawa (r)₄) = 10 cm = 10/100 mita = 1/10 mita

An tambaya: Lokacin rashin ƙarfin tsarin

Amsa:

Ni = m₁ r₁² +m₂ r₂² +m₃ r₃² +m₄ r₄²

I = (1/4)(0)² + (1/2)(0)² + (1/4)(1/10)² + (1/4)(1/10)²

I = 0 + 0 + (1/4)(1/100) + (1/4)(1/100)

I = 1/400 + 1/400

I = 2/400

I = 1/200 kg.m²

Amsar da ta dace ita ce A.

24. An haɗa ƙwallaye biyu da aka ɗauka a matsayin barbashi ta hanyar igiyar waya kamar yadda aka nuna a hoton. Idan yawan ƙwallayen P da Q sune gram 600 da gram 400 bi da bi, to lokacin inertia na tsarin ƙwallayen biyu game da axis AB shine…

A. 0,008 kg m²
B. 0,076 kg m²
C. 0,124 kg m²
D. 0,170 kg m²
E. 0,760 kg m²

Tattaunawa
An sani :
Axis na juyawa shine AB.
m_p = gram 600 = 0,6 kg, m_q = gram 400 = 0,4 kg
r_p = 20 cm = 0,2 m, r_q = 50 cm = 0,5 m
An tambaya : Lokacin rashin ƙarfin tsarin?
Jawab :
Ni = m_p r_p² +m_q r_q²
I = (0,6 kg)(0,2 m)² + (0,4 kg)(0,5 m)²
I = (0,6 kg)(0,04 m²) + (0,4 kg)(0,25 m²)
I = 0,024 kg m² + 0,1 kg m²
I = 0,124 kg m²
Amsar da ta dace ita ce C.

25. Abubuwa biyu masu tauri, silinda da kuma ƙwallo, suna kan wani wuri mai faɗi. Dukansu abubuwa suna jan su da ƙarfi iri ɗaya, inda aka taɓa su a tsakiyarsu, kuma suna da saurin farko iri ɗaya. Wanne daga cikin waɗannan maganganun ne daidai?

(1) Idan radi na abubuwa biyu iri ɗaya ne kuma ƙasa tana da santsi, to saurin abubuwa biyu koyaushe iri ɗaya ne.

(2) Idan radius na ƙwallon ya fi girma kuma ƙasan ta yi kauri to saurin abubuwa biyu zai kasance iri ɗaya koyaushe.

(3) Idan radius ɗin ba iri ɗaya ba ne kuma ƙasan yana da santsi, to gudun ƙarshe na abubuwa biyu zai kasance iri ɗaya matuƙar saurin farko ya bambanta.

(4) Idan radius ɗin ba iri ɗaya ba ne kuma ƙasa tana da santsi, saurin abubuwa biyu zai kasance daban-daban koyaushe.

Tattaunawa

Lokacin inertia na silinda mai ƙarfi = 1/2 mR²

Lokacin inertia na ƙwallon da tauri = 2/5 mR²

A ce tarin abubuwan biyu iri ɗaya ne kuma radi na abubuwan biyu iri ɗaya ne, to lokacin inertia na silinda ya fi girma (1/2) yayin da lokacin inertia na ƙwallo mai ƙarfi ya fi ƙanƙanta (2/5). Don haka, saurin ƙwallo ya fi girma saboda lokacin inertia ɗinsa ya fi ƙanƙanta.

Idan radius na sphere ya fi girma, lokacin inertia na sphere yana ƙaruwa kuma yana iya daidaita lokacin inertia na sphere. Saboda haka, saurin sphere da cylinder na iya zama iri ɗaya.

Bayanin da ya dace (2)

 

Tushen tambaya:

Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a