Phoenix Business Daily

What is the moment of inertia of this triangle about an axis perpendicular to its plane?

Three 0.10kg balls are connected by light rods to form an equilateral triangle with a side length of 0.40m . What is the moment of inertia of this triangle about an axis perpendicular to its plane and passing through one of the balls? Thank you, now I understand how to do the problem

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1. the moment of inertia due to point masses, as are these balls, is given by sum(mr^2) where m is the mass of each object and r is its distance from the rotation axis each ball has mass of 0.10kg; one ball is on the rotation axis, so its value of r is zero, and the other two are 0.40m from the axis so the moment of inertia is I = 0.10kg x 0^2 + 0.1kgx(0.4m)^2+0.1kgx(0.4m)^2 I=0.032kgm^2