tekis shakilning enersiya momenti.
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Scroll down 👇![tekis shakilning enersiya momenti. reja: 1. statik moment. og`irlik markazi. a) tekis egri chiziqning og`irlik markazi b) tekis shaklning og`irlik enersiya momenti 2. inersiya momenti yuqorida, to`g`ri chiziqli yo`lda o`zgaruvchi kuchning bajargan ishini, o`zgaruvchi zichlikka ega bo`lgan sterjen (tayoqcha) massasini hisoblash masalalarini qarab, ularni aniq integral orqali ifodalagan edik. ya`ni o`zgaruvchi f(x) kuchning [a;b] kesmadan iborat yo`lda bajargan ishi uchun formulani, o`zgaruvchi ((x) zichlikka ega bo`lgan [a;b] kesmadan iborat sterjen massasi uchun esa (*) formulani oldik. endi mexanikaning ba`zi bir masalalarini aniq integral yordamida yechish misollarini keltiramiz. 1. statik moment. og`irlik markazi. aytaylik, n ta a1,a2,..,an moddiy nuqtalarning qandaydir sistemasi berilgan bo`lib, ularning massalari mos ravishda m1,m2,..., mn bo`lsin. bu ai(xi;yi) (i=1,2,...,n) har bir nuqtadan koordinat o`qlarigach bo`lgan masofalarining uning massasiga ko`paytmasi ximi, yimi, i=1,2,..,n mi massaning mos ravishda oy va ox o`qlariga nisbatan statik momentlari deyiladi. agar bu moddiy nuqtalarning barchasi bitta tekislikka joylashgan bo`lsa, bu tekislikda kiritilgan dekart …](/media/previews/pages/preview_X499xoW.webp "tekis shakilning enersiya momenti. - Page 1")
![sifatida a nuqtasidan boshlab hisoblangan qaralayotgan c(x;y) nuqtasigacha bo`lgan yoy bo`lagining uzunligi undan tashqari, bu nuqtadagi yoy zichligi ((t) dan iborat deb qabul qilingan bo`lsin. agar ab yoy uzunligini s bilan belgilasak, t( [0;t] bo`lishi ravshandir. [0;t] kesmani (ya`ni ab yoyni) ixtiyoriy tanlangan 0=t0 restart; r>0: > f:=x->sqrt(r^2-x^2); > dif1:=diff(f(x),x); > l:=int(sqrt(1+dif1^2),x=0..r); > mx:=int(f(x)*sqrt(1+dif1^2),x=0..r); warning, unable to determine if -r is between 0 and r; try to use assumptions or set _envallsolutions to true > my:=int(x*sqrt(1+dif1^2),x=0..r); > xc:=my/l; yc:=mx/l; > r:=2:xc;yc; 36-misol. x=a(t–sint), y=a(1–cost)(a>0) sikloida bitta arkining (0(t(2() uzunligini og`irlik markazining koordinanalarini hisoblang (14- rasm). yechish. 1)egri chiziq grafigi: > with(plots): > plot([1*(t-sin(t)), 1*(1-cos(t)), t=0..2*pi]); 2) yoy uzunligini hisoblaymiz: x`=a(1-cost), y`=asint = 3) , 37-misol. qutb koordinatalar tekisligida berilgan (=a(1+cos() kordioida (16-rasm) yoyi bo`lagining og`irlik markazining koordinatalarini hisoblang. yechish: kordioida bo`lagining massasi oy o`qqa nisbatan statik momenti: = ox o`qqa nisbatan statik momenti: > restart; > with(plots):with(student[calculus1]): > plot([2*(1+cos(t)),t,t=0..2*pi],coords=polar,thickness=2); > …](/media/previews/pages/preview_2u1kzMK.webp "tekis shakilning enersiya momenti. - Page 2")
![0..2); > my:=int(x*(2*x-x^2),x=0..2)=int(x*(2*x-x^2),x=0..2); > my:=int(x*(2*x-x^2),x=0..2); > my/s; mx/s; 1 41-misol. x2+y2=9 aylana va 4x2+9y2=36 ellips bilan chegaralangan birjinsli shaklning 1-chorakdagi bo`lagining og`irlik markazi topilsin (26- rasmga qarang). masalada ko`rsatilgan shakilning og`irlimk markazi nuqtada. bu misoldan ko`rinadiki, og`irlik markazi bu shaklning simmetriya o`qidadir. > restart;with(plots): > implicitplot([x^2+y^2=9,4*x^2+9*y^2=36], x=0..3, y=0..3, color= [blue,red], thickness=2); 26- rasm > f1:=x->2*sqrt(9-x^2)/3;f2:=x->sqrt(9-x^2); > my:=int(x*(f2(x)-f1(x)),x=0..3); > mx:= int((f2(x)^2-f1(x)^2)/2,x=0..3); > s:=int(f2(x)-f1(x),x=0..3); > xc:=my/s; yc:=mx/s; egri chiziq tenglamalari parametrik ko`rinishda bo`lganda yuqoridagi masalani yechamiz. > x2:=t->3*cos(t);y2:=t->3*sin(t); > x1:=t->3*cos(t);y1:=t->2*sin(t); > s1:=int(x1(t)*diff(y1(t),t),t=0..pi/2); > s2:=int(x2(t)*diff(y2(t),t),t=0..pi/2); > s:=s2-s1; > my:=int((x1(t)*y1(t)*diff(x1(t),t)-x2(t)*y2(t)*diff(x2(t),t)), t=0..pi/2); > mx:=int((y1(t)^2*diff(x1(t),t)-y2(t)^2*diff(x2(t),t))/2, t=0..pi/2); > xc:=my/s; yc:=mx/s; og`irlimk markazi c(1;0.4) nuqtada. bu misoldan ko`rinadiki, og`irlik markazi bu shaklning simmetriya o`qidadir. 25-rasm. 26-rasm. bu umumiy xossa bo`lib, agar birjinsli shakl biror o`qqa nisbatan simmetrik bo`lsa, uning og`irlik markazi shu simmetriya o`qida yotadi. (33) formulalarning ikkinchisini ko`rinishga keltirish qiyin emas. bundan ko`rinadiki, tenglikning o`ng tomoni (x([a;b] 0( f1(x)(f2(x) faraz asosida tekis shaklni ox o`qi atrofida aylantirishdan …](/media/previews/pages/preview_Nd8jCt4.webp "tekis shakilning enersiya momenti. - Page 4")
![mi shakl yuzini og`irlik markazi aylanishidan hosil bo`lgan aylana uzunligiga ko`paytmasiga tengdir: v=2(ys. 27- rasm 42-misol. radiusi r ga teng bo`lgan birjinsli yarim doiraning og`irlik markazini toping (27- rasmga qarang). yechish. agar yarim doirani 27- rasmdagidek joylashtirsak, u oy o`qqa nisbatan simmetrikdir. demak og`irlik markazi shu o`qda yotadi: x=0. guldenning ikkinchi teoremasi asosida bundan demak, og`irlik markazi c nuqtada. 40-misol. qutb koordinatalar tekisligida berilgan (=a(1+cos() kordioida bilan figura og`irlik markazining dekart koordinatalarini hisoblang. yechish. suratdagi integralni hisoblaymiz: bunda birinchi va ikkinchi qo`shiluvchilarnin integrali ga bogliq funktsiya bo`ganligi uchun nolga teng. qolgan integralini topamiz: = maxrajdagi integralni topamiz: bundan figura ox o`qqa simmetrikligidan bo`ladi. demak: c . > restart;with(student[calculus1]): > p:=phi->a*(1+cos(phi)); > s:=int(p(phi)^2,phi=0..2*pi)/2; > mx:=int(p(phi)^3*sin(phi),phi=0..2*pi)/3; > my:=int(p(phi)^3*cos(phi),phi=0..2*pi)/3; > xc:=my/s; yc:=mx/s; 2. inersiya momenti. agar a1, a2,.., an nuqtalarga joylashgan va massalari mos ravishda m1,m2,…, mn bo`lgan n ta moddiy nuqtalarning sistemasi berilgan bo`lib, nuqtalar ai(xi;yi) koordinatalarga ega bo`lsa, bu moddiy nuqtalar …](/media/previews/pages/preview_g9bkgke.webp "tekis shakilning enersiya momenti. - Page 5")
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tekis shakilning enersiya momenti. reja: 1. statik moment. og`irlik markazi. a) tekis egri chiziqning og`irlik markazi b) tekis shaklning og`irlik enersiya momenti 2. inersiya momenti yuqorida, to`g`ri chiziqli yo`lda o`zgaruvchi kuchning bajargan ishini, o`zgaruvchi zichlikka ega bo`lgan sterjen (tayoqcha) massasini hisoblash masalalarini qarab, ularni aniq integral orqali ifodalagan edik. ya`ni o`zgaruvchi f(x) kuchning [a;b] kesmadan iborat yo`lda bajargan ishi uchun formulani, o`zgaruvchi ((x) zichlikka ega bo`lgan [a;b] kesmadan iborat sterjen massasi uchun esa (*) formulani oldik. endi mexanikaning ba`zi bir masalalarini aniq integral yordamida yechish misollarini keltiramiz. 1. statik moment. og`irlik markazi. aytaylik, n ta a1,a2,..,an moddiy nuqtalarning qandaydir sistemasi berilg...
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