Ai 2 discuri:
Primul cu centrul O(x1,y1), raza R=40cm.
Al doilea cu centru O'(x2,y2), raza r=10cm.
Al doilea centru se formeaza la distanta R/2 de primul.
Nu e greu de observat ca R/2 reprezinta distanta dintre 2 puncte:
[tex]\displaystyle \frac R2=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/tex]
Pentru a afla cu cit sa deplasat centrul de greutate, trebuie mai intii sa aflam unde se afla noul centr de greutate, O"(x3,y3).
Coordonatele unui centru de greutate format din decuparea altuia, se dau de formulele:
[tex]\displaystyle x_3=\frac{x_1\times S_1-x_2\times S_2}{S_1-S_2} \\ \\
y_3=\frac{y_1\times S_1-y_2\times S_2}{S_1-S_2}[/tex]
Unde S1 reprezinta aria discului initial, iar S2 aria discului taiat:
[tex]S_1=\pi\times R^2\\ \\
S_2=\pi\times r^2[/tex]
Inlocuind, obtinem:
[tex]\displaystyle x_3=\frac{x_1\times \pi\times R^2-x_2\times \pi\times r^2}{\pi\times R^2-\pi\times r^2}=\frac{x_1\times R^2-x_2\times r^2}{R^2-r^2}~\pi~se~simplifica\\ \\
Analogic~pentru~y_3: \\ \\
y_3=\frac{y_1\times R^2-y_2\times r^2}{R^2-r^2}[/tex]
Deplasarea intre cele 2 centre o putem afla prin distanta dintre 2 puncte:
[tex]\displaystyle d=\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}\\ \\ \\
(x_3-x_1)^2=(\frac{x_1\times R^2-x_2\times r^2}{R^2-r^2}-x_1)^2=\\ \\
=(\frac{x_1\times R^2-x_2\times r^2-x_1\times R^2+x_1\times r^2}{R^2-r^2})^2=(\frac{x_1\times r^2-x_2\times r^2}{R^2-r^2})^2=\\ \\
(\frac{r^2\times(x_1-x_2)}{R^2-r^2})^2=\frac{r^4\times(x_1-x_2)^2}{(R^2-r^2)^2}\\ \\ \\
Analogic~pentru~(y_3-y_1)^2: \\ \\
(y_3-y_1)^2=\frac{r^4\times(y_1-y_2)^2}{(R^2-r^2)^2}\\ \\ \\
Inlocuind~in~formula,~obtinem:\\ \\
[/tex]
[tex]\displaystyle d=\sqrt{\frac{r^4\times(x_1-x_2)^2}{(R^2-r^2)^2}+\frac{r^4\times(y_1-y_2)^2}{(R^2-r^2)^2}}=\\ \\
=\sqrt{\frac{r^4\times(x_1-x_2)^2+r^4\times(y_1-y_2)^2}{(R^2-r^2)^2}}=\\ \\
\sqrt{\frac{r^4\times[(x_1-x_2)^2+(y_1-y_2)^2]}{(R^2-r^2)^2}}=\\ \\
\frac{r^2}{R^2-r^2}\times\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ \\ \\
Daca~ne~uitam~la~prima~formula~scrisa~obtinem~formula~finala:\\ \\ \\
d=\frac{r^2}{R^2-r^2}\times\frac R2=\frac{r^2\times R}{2\times(R^2-r^2)}[/tex]
[tex]\displaystyle Calculam: \\ \\
d=\frac{10^2\times 40}{2\times(40^2-10^2)}=\frac{4000}{3000}=\frac 43[/tex]
