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 2.5 R 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 can be 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 = 3.5R (R + 2.5R). So, I = (2/5)MR^2 + M(3.5R)^2 = (2/5)MR^2 + 12.25MR^2 = (82/5)MR^2.

A Solid Uniform Sphere of Radius R and Mass M