I'm in a multivariable calculus course and we are currently going over center of mass using integrals. I am having trouble with these two problems.
1)Given a, b > 0, determine the center of mass of a homogeneous triangle with vertices (0, 0), (1, 0), and (a, b). Show that it lies at the intersection of the medians.
2)Determine the center of mass of the homogeneous sector 0 ≤ θ ≤ π/6, 0 ≤ r ≤ 1. Determine the moment of inertia of the sector for rotations about the axis passing though the center of mass and perpendicular to the plane of the sector.
I am lost on how to prove it lies on the intersection in 1.
For 2) I have the center of mass but then I kinda get lost in all the calculations in the moment of inertia calculation.
Any help is appreciated! Thanks!
Welcome to the forum.
For 1, I'd divide the triangle into horizontal parallel strips. (see diagram below)
I assume you can use the property 'centre of mass of a thin strip' lies at its centre. If not you'd have to prove that first.
So for each strip the C of M lies on the median at a vertical distance from the x axis, y, that is proportional to its length, L.
(That comes straight from a Euclidean theorem)
So you can integrate in the y direction.
ps. I'll come back to the Moment of Inertia later.
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