Answer to Question #109134 in Calculus for Chinmoy Kumar Bera

Question

any triangular area ABC find the moment of inertia about a perpendicular to its plane through A

Answer 1

Let a triangle be given MNK,MN=b,NK=a,MK=c. Define a coordinate system oxyz the point o coincides with point M, the axis ox is along MK, oy is perpendicular to MK, oz is the right triple with ox and oy. The task requires finding the moment of inertia about the oz axis.

Since the body is flat, I_z = I_x + I_y. Calculate I_x and I_y:

Denote h is the height of the triangle, omitted from the vertex N, c₁ is the length of the segment of side c from point M to the base of the height, S is the area of ​​the triangle. Then

:$I_x = \int y^2\ dm =\\ = \frac{m}{S}\left(\int_{0}^{\frac{hx}{c_1}}\int_{0}^{c_1}y^2\ dydx\ +\int_{0}^{\frac{h(c-x)}{c-c_1}}\int_{c_1}^{c}y^2\ dydx\right) = \frac{mh^2}{6}$

Similarly, we get $I = \frac{m}{6}(cc_1+c_1^2+c^2)$

$S = \frac{hc}{2},h^2=b^2-c_1^2, c_1 = \frac{b^2+c^2-a^2}{2c}$

$I_z=\frac{m}{12}(3b^2+3c^2-a^2)$