Question

1. Consider 3 equal masses having values m = 50g fixed to the vertices of an equilateral triangle with sides
of length a = 20cm as shown below. The masses are connected by thin rods of negligible mass.
m
60°
(a) (3 points) Find the center of mass of the system.
(b) (3 points) Find the moment of inertia for rotations of the system about an axis perpendicular to
the plane of the triangle and passing through point 0.
(c) (3 points) Find the moment of inertia for rotations of the system about an axis perpendicular to the
plane of the triangle and passing through the center or mass. Hint: Use the parallel axis theorem

Answer 1

Answer:

Explanation:

1a) x = 50(0) + 50(10) + 50(20) / 3(50) = 10 cm

    y = 50(0) + 50(0) + 50(20sin60) / 3(50) = (10/3)√3 cm

(10, (10/3)√3)

1b) I = 0.050(0²) + 0.050(.20²) + 0.050(.20²) = 0.004 kg•m²

1c) distance of each mass from the CM is

d² = 10² + ((10/3)√3)²

d² = 100 + ((100/9)3)

d² = 100 + 100/3

d² = 400/3 cm² = 0.04/3 m²

I = 3(0.050)(0.04/3) = 0.002 kg•m²