jismlarning og`irlik markazini aniqlash usullari

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1526622893_71625.doc v y y v x x k k c k k c u u å = å = ; r m r m 1 k v r m r m 1 r m r n r n r m 1 k x k y k x k y k v r v å k x r v å r v å k x 1 v 1 x 2 v 2 x n v n x 1 v 1 x 2 v 2 x n v n x c z v 1 k v å k u k k k k n m n m 1 = 0 ... ... 2 2 1 1 2 2 1 1 = - - - - + + + = å n n n n k k z z z z z z z u u u u u u u k k …
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alardan o`qigacha bo`lgan masofalar teng; embed equation.3 = embed equation.3 demak, bu nuqtalarning va koordinatalari o`zaro tng ishoralari esa, teskari bo`ladi. u holda har bir , , zk koordinatalar bilan aniqlanadigan xajmli bo`lakchaga mos keladi. shu sababli = 0 va yk= 0 tng bo`ladi. embed equation.3 = embed equation.3 + embed equation.3 +…….. + embed equation.3 - embed equation.3 - embed equation.3 -…….- embed equation.3 = 0 shuning uchun xc = 0 va us = 0 jismning og`irlik markazi z o`qida yetadi va uning bu o`qdagi holati bitta koordinata bilan aniqlanadi: = (94’) 2 – teorema: agar jism simmetriya tekisligiga ega bo`lsa, jismning og`irlik markazi shu simmetriya tekisligida yotadi (88- rasm). buni isbot qilish uchun simmetriya tekisligi orqali oxy tekslikni o`tkazamiz bu tekislikka perpendikulyar qilib z o`qini yo’naltiramiz. jismdan oxy tekisligiga nisbatan simmetrik joylashgan ikkita mk va mk1 nuqtalarni olamiz. bu nuqtalarning atrofidan elementar xajmlarni ajratib olamiz. mk va mk1 …
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uchun butun jismning va teshikning og`irlik markazlari ma’lum bo`lishi kerak. integrallash usuli. agar jism bir nyechamizta og`irlik markazlari ma’lum bo`lgan bo`lakchalarga ajratish mumkin bo`lmasa, oldin u ixtiyoriy kichik xajmlarga bo`linadi va jism uchun (95) formula quyidagi ko`rinishni oladi. va xokazo, (96) bunda xk , yk ,zk - xajm ichida yotgan biror nuqtaning koordinatalari. (96) formulalarga nolga intildirib limitga o’tsak, quyidagilarni olamiz: a) xajm og`irlik markazining koordinatalari uchun: (97) b) yuza og`irlik markazining koordinatalari uchun: ( 98) v) chiziq og`irlik markazining koordinatalari uchun: (99) uchburchak yuzasining og`irlik markazi. ixtiyoriy avd uchburchak yuzasining og`irlik markazini aniqlash uchun uchburchak yuzasini av tomoniga parallel bo`lgan to`g`ri chiziq kesmasi bilan bo`lamiz (89- rasm ). har bir bunday kesmaning og`irlik markazi uning urtasida ya’ni de medianada yotadi. demak, uchburchak yuzasining og`irlik markazi bu medianaga yotadi. xuddi shuningdek, uchburchak yuzasini ad tomoniga parallel bo`lgan to`g`ri chiziq kesmasi bilan ajratsak, bu to`g`ri chiziq kesmalarining og`irlik markazi vk medianada …
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ga teng bo`lgan elementar bo`lakcha ajratib olamiz. uning holati burchakgi bilan aniqlanadi. elementar burchakga og`irlik markazining koordinatasi ga teng (99) formulalarning birinchisiga x va de larning qiymatlarini qo`yib va butun yoyining uzunligi bo`yicha integrallaymiz: bunda l –av yoyining uzunligi ga teng. demak , aylana yoyining og`irlik markazi simmetriya o`qida, yotadi va aylana markazidan (101) 90-rasm masofada bo`ladi. bunda l burchagi radianda o`lchanadi. agar 2l = ga teng bo`lsa, yarim aylana hosil bo`ladi (90- rasm ) buni (99) formulaga qo`ysak, 91-rasm (101) formula bilan yarim aylana yoyining og`irlik markazinig koordinatasi topiladi. doira sektori yuzasining og`irlik markazi radiusi r, markaziy burchagi 2 ga teng doira sektori yuzasining og`irlik markazining aniqlash uchun x o`qni sektor yuzasining simmetriya o`qi bo`ylab yo’naltiramiz (92- rasm ). sektor yuzasining bir qancha elementar sektorlardan tashqil topgan deb karaymiz. har bir elementar sektorni balandligi r ga teng uch-burchak deb karasak, uning og`irlik markazi o nuqtadan masofada yotadi. oav doira …
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1526622893_71625.doc v y y v x x k k c k k c u u å = å = ; r m r m 1 k v r m r m 1 r m r n r n r m 1 k x k y k x k y k v r v å k x r v å r v å k x 1 v 1 x 2 v 2 x n v n x 1 v 1 x 2 v 2 x n v n x c z v 1 k v å k u k k k k n m n m 1 = 0 ... ... 2 2 1 1 2 2 1 1 = - …

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