Three 0.21 kg particles form an equilateral triangle with 0.89 m sides. The part

Question

Three 0.21 kg particles form an equilateral triangle with 0.89 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

Here's how I think you solve.

For each triangle we calculate Inertia as mr2 where m is the mass of the particle(in this case, the .21kg particles) and r is the distance from theaxis of rotation. We take the sum of each particles' masstimes the distance to the axis squared, and this gives the totalinertia.

For part a) the first particle is on the axis, so its distance iszero, which means it falls out of the equation. The second twoparticles are .89m from the axis, so their Inertia is .21(.89^2).You sum the inertias to get:
I = 0 + .21(.89^2) + .21(.89^2) = 0.332kg m2


To solve parts B and C, you do the same thing, but with eachparticle's new distance to the axis of rotation.

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