Misali na Motsin Zagaye

Misalai 11 na Tambayoyin Motsin Zagaye

1. Abu mai nauyin kilogiram 10 shinemotsi na zagaye iri ɗaya a gudun 4 ms^-1Idan radius na da'irar mita 0,5 ne, to:
(1) Mitar juyawa ita ce 4/π Hz
(2) Hanzarin tsakiya32 ms ɗinsa^-2
(3) Ƙarfin tsakiya shine 320 N
(4) Lokacin shine 4π s.
Bayanin da ya dace shine…
1. (1), (2), (3), da (4)
2. (1), (2), da (3)
3. (1), da (3) kawai
4. (2), da (4) kawai
5. (3), da (4) kawai
Tattaunawa
2. Abu yana motsawa a cikin da'ira mai fadin mita 6. Idan abu yana juyawa sau 16 cikin mintuna 2, to saurin layi na abu shine...
A. 0,8 p ms^-1
B. 1,0 na rana ms^-1
C. 1,2 p ms^-1
D. 1,4 p ms^-1
E. 1,6 p ms^-1
Tattaunawa
An san cewa:
Radius (r) = mita 6
Saurin kusurwa (ω) = juyin juya hali 16 / mintuna 2 = juyin juya hali 8 / minti = juyin juya hali 8 / daƙiƙa 60 = juyin juya hali 0,13 / daƙiƙa.
An tambaya: saurin layi (v)?
Amsa:

Tsarin dangantakar da ke tsakanin saurin layi (v) da saurin kusurwa (ω):

v = r ω = (mita 6)(juyin juya hali 0,13/daƙiƙa) = juyi 0,8 mita/daƙiƙa
Amsar da ta dace ita ce A.
Idan aka bayyana a cikin radians:
Juyin juya hali 1 = 2π radians = 2(3,14) = 6,28 radians
Gudun kusurwa = 8 (6,28) radians / daƙiƙa 60 = 50,24 radians / daƙiƙa 60 = 0,84 radians/daƙiƙa
v = r ω = (mita 6)(radian 0,84/daƙiƙa) = radian 5,04/daƙiƙa

3. Famfon propeller mai radius na 20/π cm zai iya juyawa sau 4 a cikin daƙiƙa 1. Saurin layi na ƙarshen propeller shine…
A. 3,2 ms^-1
B. 1,6 ms^-1
C. 1,3 ms^-1
D. 1,0 ms^-1
E. 0,8 ms^-1
Tattaunawa

4. Ga wani abu da ke tafiya daidai gwargwado a cikin hanyar da'ira, saurinsa na layi ya dogara da...
A. nauyi da radius na da'irar
B. taro da lokaci
C. taro da mita
D. lokaci da radius na hanyar
E. saurin kusurwa da radius na da'irar
Tattaunawa

 

5. Abu yana motsawa a cikin da'ira mai radius na 50 cm. Idan abu yana juyawa a 120 rpm, to lokacin juyawa da saurin abu sune bi da bi….
A. 0,5 s da 2π ms^-1
B. 0,5 s da 0,2π ms^-1
C. 0,5 s da π ms^-1
D. 2 s da 5π ms^-1
E. 2 s da 10π ms^-1
Tattaunawa
An san cewa:
Radius (r) = 50 cm = mita 0,5
Gudun kusurwa (ω) = 120 rpm = juyin juya hali 120 / minti 1 = juyin juya hali 120 / minti 60 = juyin juya hali 2 / daƙiƙa 1
Juyin juya hali 1 = 2π radians
Gudu na kusurwa (ω) = 2 (2π radians) / daƙiƙa 1 = 4π radians/daƙiƙa
Tambaya: Lokacin juyawa (T) da saurin layi (v)
Amsa:
Lokacin zagayowar ko lokacin (T):
Lokacin shine lokacin da abu ke ɗaukarsa kafin ya kammala juyawa ɗaya.
Abu yana yin juyi sau 2 a cikin daƙiƙa 1 = abu yana yin juyi sau 1 a cikin daƙiƙa 0,5. Don haka lokacin juyawa ko lokacin shine daƙiƙa 0,5.
Saurin abu (v):
v = r ω = (mita 0,5)(4π radians/daƙiƙa) = mita 2π/daƙiƙa
Amsar da ta dace ita ce A.

6. An ɗora tayoyi B da C a kan gatari ɗaya kuma tayoyi A tana hulɗa da tayoyi B kamar yadda aka nuna a hoton da ke gefe. Idan radius na tayoyi A yayi daidai da radius na tayoyi C = 30 cm, radius na tayoyi B = 60 cm, rabon saurin layi tsakanin tayoyi A, B, da C shine…

A. 1: 1: 1

B. 1: 2: 1

C. 1: 4: 1

D. 2: 2: 1

E. 4: 1: 2

Tattaunawa

An sani cewa :

Radius na tayoyin A (rA ₎ = 30 cm = mita 0,3

Radius ɗin tayoyin B (r _B ) = 60 cm = mita 0,6

Radius ɗin tayoyin C (r _C ) = 30 cm = mita 0,3

Ana so : Kwatanta saurin layi tsakanin tayoyin A, B, da C

Amsa :

Gudun layi na gefen tayoyin A

An haɗa ƙafafun A da ƙafafun B da juna kamar yadda yake a hoton da ke sama, saboda haka saurin kusurwa na ƙafafun A ba iri ɗaya bane da saurin kusurwa na ƙafafun B. Wannan saboda kewayen ƙafafun B ya fi na ƙafafun A girma. A lokacin lokaci ɗaya, lokacin da ƙafafun A suka motsa a cikin da'ira juyawa ɗaya (360 ^o ), ƙafafun B bai kai juyawa ɗaya ba tukuna (360 ^o ). Duk da haka, a lokacin lokaci ɗaya, nisan da gefen ƙafafun A ya yi daidai da nisan da gefen ƙafafun B ya yi. Don haka saurin layi na gefen ƙafafun A (v A ₎ iri ɗaya ne da saurin layi na gefen ƙafafun B (v _B ).

Gudun layi na gefen tayoyin A:

v _A = r _A ω _A = 0,3 ω _A

Gudun layi na gefen tayoyin B

Tayar B da tayar C suna haɗe da juna, don haka tayar B da tayar C suna juyawa tare. Idan tayar B tana motsawa a cikin da'ira juyawa ɗaya (360)^o) sannan a lokacin tazara lokaci guda, dabaran C kuma yana motsawa a cikin da'ira juyawa ɗaya (360)^oDomin suna juyawa tare, saurin kusurwa na tayoyin B (ω_B) yayi daidai da saurin kusurwa na dabaran C (ω_C) = ω. Amma Gudun layi na gefen tayoyin B (v)_B) ba iri ɗaya bane da saurin layi na gefen tayoyin C (v)_C)

Gudun layi na gefen tayoyin B:

v _B = r _B ω _B = 0,6 ω _B = 0,6 ω

Gudun layi na gefen tayoyin C:

v _C = r _C ω _C = 0,3 ω _C = 0,3 ω

Gudun layi na gefen tayoyin A (v _A ) yayi daidai da gudun layi na gefen tayoyin B (v _B )

v _A = v _B

0,3 ω _A = 0,6 ω

ω _A = 0,6 ω / 0,3

ω _A = 2 ω

Gudun layi na gefen tayoyin A

v _A = 0,3 ω _A = 0,3 ( 2 ω) = 0,6 ω

Kwatanta saurin layi tsakanin tayoyin A, B, da C

v _A : v _B : v _C

0,6 ω : 0,6 ω : 0,3 ω

0,6: 0,6: 0,3

6: 6: 3

2: 2: 1

Amsar da ta dace ita ce D.

7. An haɗa tayoyi uku kamar yadda aka nuna a hoton da ke ƙasa.

Idan R _A = 10 cm, R _B = 4 cm da R _C = 40 cm, to rabon saurin kusurwa na tayoyin A da tayoyin C shine…

A. 1: 1

B. 1: 2

C. 1: 4

D. 2: 1

E. 4: 1

Tattaunawa

An sani cewa :

Radius na tayoyin A (rA ₎ = 10 cm

Radius ɗin tayoyin B (r _B ) = 4 cm

Radius ɗin tayoyin C (r _C ) = 40 cm

Tambaya : Rabon saurin kusurwa na ƙafa A da ƙafa C

Amsa :

Gudun kusurwa na ƙafafun A da C

Da'irar dabaran A ta fi girma fiye da da'irar dabaran C. Lokacin da dabaran C ta motsa a cikin da'ira juyin juya hali ɗaya (360 ^o ), a cikin tazara lokaci ɗaya dabaran A bai kai juyin juya hali ɗaya ba (360 ^o ). Don haka, saurin kusurwar dabaran A ba iri ɗaya bane da saurin kusurwar dabaran C.

Duk da haka, ana haɗa ƙafafun A da C da igiya ko sarka. Saboda suna da alaƙa da juna, a cikin tazara ɗaya, nisan da gefen ƙafafun A ke tafiya daidai yake da nisan da gefen ƙafafun C ke tafiya. Don haka saurin layi na gefen ƙafafun C (v _C ) iri ɗaya ne da saurin layi na gefen ƙafafun A (v _A ).

v _A = v _C

r _A ω _A = r _C ω _C

10 ω _A = 40 ω _C

ω _A / ω _C = 40 / 10

ω _A / ω _C = 4 / 1

Amsar da ta dace ita ce E.

8.

Tayoyi uku A, B da C suna da alaƙa da juna kamar yadda aka nuna a cikin hoton. Idan radius na ƙafafun A, B da C sun kasance 20 cm, 8 cm da 4 cm bi da bi, kuma ƙafafun B yana juyawa a kusurwar gudu na 10 rad.s^-1, sannan dabaran C yana juyawa a saurin kusurwa na...

A. 80 rad.s^-1
B. 50 rad.s^-1
C. 40 rad.s^-1
D. 20 rad.s^-1
E. 10 rad.s^-1

Tattaunawa
An sani :
Radius na tayoyin A (r)_A) = 20 cm = mita 0,2
Radius na tayoyin B (r)_B) = 8 cm = mita 0,08
Radius na tayoyin C (r)_C) = 4 cm = mita 0,04
Gudun kusurwa na tayoyin B (ω)_B) = 10 rad/s
An tambaya : Gudun kusurwa na tayoyin C (ω_C)
Jawab :

Gudun kusurwa da saurin layi na gefen tayoyin A
Tayoyi A da ƙafafun B suna haɗe da juna, don haka tayoyi A da ƙafafun B suna juyawa tare. Idan tayoyi B suna motsawa a cikin da'ira juyawa ɗaya (360)^o) sannan a lokacin tazara lokaci ɗaya, dabaran A kuma yana motsawa a cikin da'ira juyawa ɗaya (360)^oDomin suna juyawa tare a lokacin saurin kusurwa dabaran A (ω)_A) yayi daidai da saurin kusurwa na tayoyin B (ω_B).
Gudun kusurwa na tayoyin A :
ω_A = ω_B = radian 10/daƙiƙa
Gudun layi na gefen tayoyin A :
Ana ƙididdige girman saurin layi na gefen tayoyin A ta amfani da dabarar da ke nuna alaƙar da ke tsakanin saurin layi da saurin kusurwa, v = r ω.
v_A =r_A ω_A = (0,2 m)(10 rad/s) = 2 m/s

Gudun kusurwa da saurin layi na gefen tayoyin C
Da'irar dabaran A ta fi da'irar dabaran C girma. Lokacin da dabaran C ta motsa a cikin da'ira juyin juya hali ɗaya (360)^o), a lokacin tazara lokaci ɗaya, dabaran A bai kammala juyawa ɗaya ba tukuna (360^oSaboda haka, saurin kusurwar dabaran A ba iri ɗaya bane da saurin kusurwar dabaran C.
Duk da haka, ana haɗa ƙafafun A da C da igiya ko sarka. Saboda an haɗa su, a cikin tazara ɗaya, nisan da gefen tayoyin A ya yi daidai da nisan da gefen tayoyin C ya yi. Don haka saurin layi na gefen tayoyin C (v)_C) yayi daidai da saurin layi na gefen tayoyin A (v_A).
Gudun layi na gefen tayoyin C :
v_C = v_A = 2m/s
Gudun kusurwa na dabaran C :
v_C =r_C ω_C
ω_C = v_C /r_C = 2 / 0,04 = 50 radians/daƙiƙa = 50 rad.s^-1.
Amsar da ta dace ita ce B.

9.

Tsarin ƙafafun R-spoke_A = 2 cm; R_B = 4 cm da R_C = 10 cm an haɗa shi kamar yadda aka nuna a hoton. Tayar B tana juyawa a juyawa 60 a minti ɗaya, don haka saurin layi na tayar C shine…

A. 8π cm.s^-1
B. 12 cms.s^-1
C. 12π cm.s^-1
D. 24 cms.s^-1
E. 24π cm.s^-1

Tattaunawa
An sani :
Radius na tayoyin A (r)_A= 2 cm
Radius na tayoyin B (r)_B= 4 cm
Radius na tayoyin C (r)_C= 10 cm
Gudun kusurwa na tayoyin B (ω)_B) = Juyi 60/minti = Juyi 60/daƙiƙa 60 = Juyi 1/daƙiƙa = 1(2π radian)/daƙiƙa = 2π rad/s
An tambaya : Gudun layi na tayoyin C (v)_C)
Jawab :
Gudun layi na gefen tayoyin B
Gudun layi na gefen tayoyin B:
v_B =r_B ω_B = (4 cm)(2 π rad/s) = 8 π cm/s
Gudun layi na gefen tayoyin A
Ana haɗa tayoyin A da ƙafafun B ta hanyar igiya don haka saurin layi na gefen tayoyin A (v)_A) yayi daidai da saurin layi na gefen tayoyin B (v_B).
v_A = v_B = 8π cm/s
Gudun layi na gefen tayoyin C
Ana haɗa ƙafafun C da ƙafafun A ta hanyar igiya don haka saurin layi na gefen ƙafafun C shine (v)_C) yayi daidai da saurin layi na gefen tayoyin A (v_A).
v_C = v_A = v_B = 8π cm/s
Amsar da ta dace ita ce A.

10.
Kula da waɗannan hanyoyin haɗin ƙafafun! Radius ɗin ƙafafun shine R_A = 25 cm, R_B = 15 cm, R_C = 40 cm kuma tayoyin C suna juyawa a saurin juyawa na juyawa 60 a minti daya. Gudun kusurwa na tayoyin A shine…
A. 2,5π rad.s^-1
B. 3π rad.s^-1
C. 3,2π rad.s^-1
D. 3,5π rad.s^-1
E. 3,8π rad.s^-1
Tattaunawa
An sani :
Radius na tayoyin A (r)_A) = 25 cm = mita 0,25
Radius na tayoyin B (r)_B) = 15 cm = mita 0,15
Radius na tayoyin C (r)_C) = 40 cm = mita 0,4
Gudun kusurwa na dabaran C (ω)_C) = Juyi 60/minti = Juyi 60/daƙiƙa 60 = Juyi 1/daƙiƙa = 1(2π radian)/daƙiƙa = 2π rad/s
An tambaya : Gudun kusurwa na tayoyin A (ω_A)
Jawab :
Gudun layi na gefen tayoyin C
Gudun layi na gefen tayoyin C:
v_C =r_C ω_C = (0,4 m)(2π rad/s) = 0,8π m/s
Gudun layi na gefen tayoyin B
An haɗa ƙafafun C da ƙafafun B ta hanyar igiya don haka saurin layi na gefen ƙafafun C shine (v)_C) yayi daidai da saurin layi na gefen tayoyin B (v_B).
v_B = v_C = 0,8π m/s
Gudun kusurwa da saurin layi na gefen tayoyin A
An haɗa ƙafafun A da ƙafafun B da juna kamar yadda aka nuna a hoton da ke sama, saboda haka saurin kusurwar ƙafafun A ba iri ɗaya bane da saurin kusurwar ƙafafun B. Wannan saboda kewayen ƙafafun A ya fi na ƙafafun B girma. A lokacin lokaci ɗaya, lokacin da ƙafafun A suka motsa a cikin da'ira juyin juya hali ɗaya (360)^o), dabaran B bai kai ga juyawa ɗaya ba tukuna (360^oDuk da haka, a cikin tazara ɗaya, nisan da gefen tayoyin A ya yi daidai da nisan da gefen tayoyin B ya yi. Don haka saurin layi na gefen tayoyin A (v_A) yayi daidai da saurin layi na gefen tayoyin B (v_B).
Gudun layi na gefen tayoyin A
v_A = v_B = v_C = 0,8π m/s
Gudun kusurwa na tayoyin A
v_A =r_A ω_A
ω_A = v_A /r_A = 0,8π / 0,25 = 3,2π rad/s
Amsar da ta dace ita ce C.

11. Ana ɗaure dutse mai nauyin m a ƙarshen igiya mai tsawon l sannan a juya shi da saurin kusurwa ω har sai hanyar dutsen ta yi zagaye a cikin jirgin kwance kuma igiyar ta samar da kusurwa θ zuwa tsaye (duba hoto). Gudun kusurwar dutsen shine...

Tattaunawa

Amsar da ta dace ita ce C.