Misalai 7 na tambayoyin hanzari na tsakiya
1. Ana ɗaure dutse a kan igiya sannan a juya shi a kwance har sai dutsen ya juya kusurwar 360°.o a kowace daƙiƙa 1. Idan tsawon igiyar ya kai santimita 20, a ƙayyade girman hanzarin centripetal na abin!
Tattaunawa
An sani :
Gudun kusurwa (ω) = 360o/daki = juyawa 1 / dakika = 6,28 radians/dakika
Tsawon igiya = radius (r) na hanyar da'ira = 20 cm = mita 0,2
An tambaya : hanzarin tsakiya (as)
Jawab :
2. Tayar da ke da radius na santimita 30 tana juyawa a gudun kusurwa na 180 rpm. Kayyade hanzarin tsakiya na wani wuri a gefen tayar!
Tattaunawa
An sani :
Radius (r) na tayoyin = 30 cm = mita 0,3
Gudun kusurwa (ω) = juyin juya hali 180 / daƙiƙa 60 = juyin juya hali 3 / daƙiƙa = (3)(radian 6,28) / daƙiƙa = radian 18,84/daƙiƙa
An tambaya : hanzarin tsakiya (as) a r = mita 0,3
Jawab :
3. Mai tsere yana tuka mota a kusa da lanƙwasa mai kusurwa huɗu tare da radius na lanƙwasa na mita 50. Idan gudun motar ya kai kilomita 72/awa, menene saurin centripetal na motar?
Tattaunawa
An sani :
Radius (r) = mita 50
Gudun (v) = 72 km/awa = (72)(mita 1000) / daƙiƙa 3600 = mita 20/daƙiƙa
An tambaya : girman hanzarin centripetal (a)s)
Jawab :
Girman saurin motar a tsakiyar mota:
as = v2 / r = 202 / 50 = 400 / 50 = 8 m/s2
4. Mota tana tafiya a kan lanƙwasa tare da saurin gudu na radial na 10 m/s2, inda wannan shine matsakaicin saurin centripetal da motar zata iya cimmawa ba tare da ta tsallake layin dogo ba. Idan gudun motar ya kai kilomita 108/h, menene radius na lanƙwasa na lanƙwasa don motar ba ta zame ba?
Tattaunawa
An sani :
Hanzarin tsakiya (a)s) = 10 m/s2
Saurin mota (v) = 108 km/awa = (108)(1000) / 3600 = mita 30/daƙiƙa
An tambaya : radius (r)
Jawab :
Radius na lanƙwasa na lanƙwasa:
r= v2 / Zuwas
r = 302 / 10 = 900 / 10 = mita 90
5. Tambayoyin Jarrabawar Ƙasa na 2003/2004
Jadawalin da ke nuna alaƙar da ke tsakanin hanzarin tsakiya (a s ) da saurin layi (v) a cikin motsi mai zagaye iri ɗaya...
Tattaunawa
Tsarin hanzarin tsakiya:
Bayani: a s = hanzarin centripetal, v = saurin layi, r = nisa daga axis na juyawa.
Mun binciki dangantakar da ke tsakanin hanzarin tsakiya (a s ) da saurin layi (v) don haka nisan da ke tsakanin axis na juyawa (r) ya kasance mai dorewa. A ce r = 1.
Darajar v tana ƙaruwa a layi, ƙimar kamar tana ƙaruwa sosai.
Amsar da ta dace ita ce A.
6. Tambayoyin Jarrabawar Ƙasa don SMA/MA Physics U-ZC-2013/2014 Lamba ta 4
Ana jujjuya marmara a cikin kwano mai zagaye mai diamita na mita 1. Idan marmarar tana juyawa a gefen kwano a saurin kusurwa na 50 rpm akai-akai, to saurin layi da haɓaka centripetal na marmarar sune….
A. 50 ms -1 da 2500 ms -2
B. 5π ms -1 da 25π 2 ms -2
C. 5π/3 ms -1 da 25π 2/18 ms -2
D. 5π/6 ms -1 da 25π 2/18 ms -2
E. 5π/12 ms -1 da 25π 2/36 ms -2
Tattaunawa
An san cewa:
Diamita na da'ira (D) = mita 1
Radius na da'ira (r) = mita 0,5
Gudun kusurwa (ω) = 50 rpm = juyin juya hali 50 / minti 1
Juyin juya hali 1 = 2π radians
Juyin juya hali 50 = 50 (2π radians) = 100π radians
Minti 1 = daƙiƙa 60
Gudun kusurwa (ω) = 100π radians / daƙiƙa 60 = (10π/6) radians/daƙiƙa
Tambaya: Saurin layi (v) da hanzarin centripetal (a s )
Amsa:
Gudun layi (v):
v = r ω = (0,5) (10π/6) = 5π/6 m/s
Hanzarin tsakiya (a)s):
a s = v 2 /r = (5π/6) 2 : 0,5 = 25π 2/36 : 0,5 = (25π 2/36 ) (1/ 0,5)
a s = (25π 2/18 ) m/s 2
Amsar da ta dace ita ce D.
7. Tambayoyin Jarrabawar Ƙasa na 2007/2008 P4 No. 6
Abu yana motsawa a gudun da ba ya canzawa v tare da hanyar da'ira tare da radius R tare da hanzarin centripetal (a s ). Domin saurin centripetal ya ninka na asali sau biyu, to...
Ana yin A. v sau 4 kuma ana yin R sau 2 kuma
Ana yin B. v sau 2 kuma ana yin R sau 4 kuma
Ana yin C. v sau 2 kuma ana yin R sau 2 kuma
D. v ya rage kuma R ya ninka sau biyu
E. v ya sake ninkawa kuma R ya kasance
Tattaunawa
