Question
An equilateral triangular plate has a mass of M, the side length is R. What is the center of-mass moment of Inertia, Icm, Expressed in terms of M and R? (The rotational axis is z axis, coming out of the paper perpendicularly)
Hint:
Start from a bar, located at from y to y+dy (as indicated), with a mass of dM, and use
parallel axis theorem to derive this bar’s moment of inertia dIcm relative to the center of mass. Both dM and L (the length of the bar) should be defined using the variable y. You also need to define the range of y for the integral that will be used in the end. Your final result of Icm should only be in the form of MR2 , multiplied by a constant number. This constant number is what you need to calculate.
Explanation / Answer
height of triangle is h = ¼ L3 if triangle is one of its edge on below position, so the height of center of mass is h. I = r² dA x = yL/h = dA = x dy = (yL/h) dy moment inertia with respect to X-axis is Ix = y² dA = y² (yL/h) dy = (L/h) y³ dy Ix = (L/h) (¼) y which y from 0 to h Ix = (L/h) (h - 0) = ¼ Lh³ according to parallel axis theorem, calculate moment inertia w.r.t X-axis through the center of mass is, Ix = Ixo + Ay² ¼ Lh³ = Ixo + (½ Lh) (h)² Ixo = (1/36) Lh³ Let's we calculate moment inertia with respect to Y-axis through the center of mass is, Iy = x² dA dA = y dx y = -(x - h) Iyo = x² dA = 2 x² (-(x - h)) dx from x = 0 to x = ½ L Iyo = 2 (-x³ + x²h) dx = -½ x + x³h from x = 0 to x = ½ L Iyo = (-1/32) L + (1/12) L³h notify that Y-axis through the center of mass. according to Pythagoras theorem, r² = x² + y² r² dA = x² dA + y² dA Icm = Iyo + Ixo Icm = (-1/32) L + (1/12) L³h + (1/36) Lh³ but h = ¼ L3 Icm = L(173 - 24)/768 moment of inertia of an equilateral triangular plate of length L about its centre of mass is Icm = L(173 - 24)/768