The game of Jacks is played with pieces like those shown to the right, consisting of four small spheres at the ends of two perpendicular rods of length L and mass M. The piece is spun as shown about an axis through the third perpendicular rod of length L and mass M. Assume that the mass of each sphere is M/4, and that the rotational inertia of a rod about an axis perpendicular to the center of the rod is ML2/12. The rotational inertia of the Jacks piece is closest to

A) 5/12 ML2 B) 1/3 ML2 C) 1/2 ML2 D) 13/12 ML2

To determine the rotational inertia of the Jacks piece, we need to consider the rotational inertia of each component.

First, let's consider the rotational inertia of the perpendicular rods. The rotational inertia of a rod about an axis perpendicular to its center is given by the formula ML^2/12, where M is the mass of the rod and L is its length. Since each rod has a mass of M and a length of L, the rotational inertia of each rod is ML^2/12.

Next, let's consider the rotational inertia of the small spheres. Each sphere has a mass of M/4. The rotational inertia of a sphere about an axis through its center is given by the formula 2/5 MR^2, where M is the mass of the sphere and R is its radius. Since each sphere has a mass of M/4, its rotational inertia is (2/5)(M/4)(R^2) = (1/10)MR^2.

The Jacks piece consists of four small spheres and two perpendicular rods. Therefore, the total rotational inertia of the Jacks piece is the sum of the rotational inertia of each component.

Total rotational inertia = (rotational inertia of rods) + (rotational inertia of spheres)
Total rotational inertia = (2 × (ML^2/12)) + (4 × (1/10)MR^2)
Total rotational inertia = (ML^2/6) + (2/5)MR^2

From the answer choices given, we can see that the closest value to (ML^2/6) + (2/5)MR^2 is option D) 13/12 ML^2.

Therefore, the rotational inertia of the Jacks piece is closest to 13/12 ML^2 (option D).