Three 0.70 kg particles form an equilateral triangle with 0.58 m sides. The part

Question

Three 0.70 kg particles form an equilateral triangle with 0.58 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) The perpendicular distance from the axis of rotation to either point is 0.58*sin(60) = 0.5023m

so I = 2*m*r^2 = 2*0.70kg*(0.5023m)^2 = 0.3532 kg-m^2

b) Here one of the points will be 0.5023m from the axis and the other two will be 0.58/2m away

so I = m*0.5023^2 + 2*m*(0.58/2)^2 = 0.70*0.5023^2 + 2*0.70*(0.58/2)^2 = 0.294 kg-m^2

c) Here all three points are the same distance from the axis = 0.5023/2 = 0.25115 m

so I = 3*m*r^2 = 3*0.70*0.25115^2 = 0.13246 kg-m^2

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