To find the moment of inertia about the y-axis for three masses in an equilateral triangle, we can use the parallel axis theorem. First, we need to find the moment of inertia of the equilateral triangle with respect to its centroid.
The centroid of an equilateral triangle is the intersection point of its medians, which is also the center of mass. The medians of an equilateral triangle are equal in length and intersect at a point that is two-thirds of the way from each vertex to the opposite side. Therefore, the distance from the centroid to each vertex is:
h = (√3/2) s
where s is the length of a side. Substituting s = 0.500 m, we get:
h = (√3/2) (0.500 m) ≈ 0.433 m
The moment of inertia of an equilateral triangle with respect to its centroid is:
I = (1/12) m s^2
Substituting m = 2.00 kg and s = 0.500 m, we get:
I = (1/12) (2.00 kg) (0.500 m)^2 = 0.0417 kg m^2
Now, we can use the parallel axis theorem to find the moment of inertia about the y-axis. The parallel axis theorem states that the moment of inertia about any axis parallel to the centroidal axis is equal to the moment of inertia about the centroidal axis plus the product of the total mass and the square of the distance between the two axes.
In this case, the y-axis is parallel to the centroidal axis and passes through the center of the equilateral triangle. The distance between the two axes is the distance from the centroid to the center, which is:
d = h/2 = (√3/4) s
Substituting s = 0.500 m, we get:
d = (√3/4) (0.500 m) ≈ 0.2165 m
The total mass of the system is:
M = 3m = 6.00 kg
Therefore, the moment of inertia about the y-axis is:
Iy = I + Md^2 = 0.0417 kg m^2 + (6.00 kg) (0.2165 m)^2 ≈ 0.573 kg m^2
So, the moment of inertia about the y-axis for three masses in an equilateral triangle with m = 2.00 kg and sides of 0.500 m is approximately 0.573 kg m^2.