A solid uniform sphere of radius R and mass M has a rotational inertia about a diameter that is given by (2/5)MR². A light string of length 3R is attached to the surface and used to suspend the sphere from the ceiling. Its rotational inertia about the point of attachment at the ceiling is: A) (2/5)MR² B) 9MR² C) 16MR² D) 47/5MR² E) (82/5)MR²

Question

Quizplus · Accepted Answer

The rotational inertia about the point of attachment is found using the parallel axis theorem: I = I_cm + Md^2, where I_cm is the rotational inertia about the center of mass, and d is the distance from the center of mass to the new axis. Here, d = 3R, so I = (2/5)MR^2 + M(3R)^2 = (2/5)MR^2 + 9MR^2 = (47/5)MR^2.

A Solid Uniform Sphere of Radius R and Mass M