Question

Three 0.29 kg particles form an equilateral triangle with 0.29 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?

Answer 1

Answer:

a.) I = 0.0389kg-m^2

b.) I = 0.0316kg-m^2

c.) I = 0.0147kg-m^2

Explanation:

a) when the axis passes through one of the particles and is parallel to the rod connecting the other two,

take the perpendicular distance from the axis of rotation to either point (r)

r = Length of equilateral triangle × sin(60)

Where

sin(60) is the angle of the equilateral triangle

Length of equilateral triangle =0.29m

0.29m × sin(60) = 0.259m = r

rotational inertia (i)

I = 2*m*r^2

(m = 0.29kg, r = 0.259m)

I = 2 × 0.29kg × (0.259m)^2

I = 0.0389kg-m^2

b) when it passes through the midpoint of one of the sides and is perpendicular to the plane of the triangle,

take one of the points as 0.258m from the axis and the other two will be 0.29/2m away

I = m × r^2 + 2 × m * (l/2)^2

(m = 0.29kg, r = 0.259m, l =0.29)

I= 0.29 × 0.259^2 + 2 × 0.29*(0.29/2)^2

I = 0.0316kg-m^2

c) take all three points to be the same distance from the axis

Therefore

r = r/2

= 0.259/2

r = 0.130m

Since it passes through the mid point of the other two sides

I = 3 × m × r^2

Where

(m = 0.29kg, r = 0.130m),

I = 3 × m × r^2

I = 3 × 0.29 × 0.130^2

I = 0.0147kg-m^2