Three 0.95 kg particles form an equilateral triangle with 0.23 m sides. The part

Question

Three 0.95 kg particles form an equilateral triangle with 0.23 m sides. The particles are connected by rods of negligible mass. What is the rotational inertia of this rigid body about (a) an axis that passes through one of the particles and is parallel to the rod connecting the other two, (b) an axis that passes through the midpoint of one of the sides and is perpendicular to the plane of the triangle, and (c) an axis that is parallel to one side of the triangle and passes through the midpoints of the other two sides?

Explanation / Answer

a). I=2*m*(r*sin60)^2 where r=0.23 m. I=0.075(kgm2). b). I=2*m*(r/2)^2+m*(r*sin60)^2=0.063(kgm2). c). I=3*m*(r*sin60/2)^2=0.028(kgm2)

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