Question

Three thin rods of length AC=BC= a,and​ AB=2^1/2a ,and each rod having mass 'm' are joined together to form a right angled triangle. Find the moment of inertia of the triangle about an axis perpendicular to its plane and passing through its center of mass.

Answer 1

\(GL = \frac{a}{3\sqrt{2}}\)

\(OG = \frac{a}{3}\)

\(GF = \frac{a}{3}\)

Moment of inertia \(I = I_1 + I_2 + I_3\)

\(I_1 = \frac{Ma^2}{12} + M\left(\frac{a}{3}\right)^2\)

\(I_2 = \frac{Ma^2}{12} + M\left(\frac{a}{3}\right)^2\)

\(I_3 = \frac{M(\sqrt{2}a)^2}{12} + M\left(\frac{a}{3\sqrt{2}}\right)^2\)

\(I = \frac{Ma^2}{12} + \frac{Ma^2}{12} + \frac{Ma^2}{9}\) \(+ \frac{Ma^2}{9} + \frac{2Ma^2}{12} + \frac{Ma^2}{18}\)

\(I = \frac{4Ma^2}{12} + \frac{5Ma^2}{18}\)

\(I = \frac{12Ma^2 + 10Ma^2}{36}\)

\(I = \frac{22Ma^2}{36}\)

Comments

Is not we required GN, GM, distance between two parallel axis..... And not OG, GF distance.... Mam....