A uniform disk (I_o = ½ mR²)of mass m and radius R is suspended from a point on its rim.The moment of inertia of the disk about an axis perpendicular to the disk through the pivot point is A)½ mR² B)mR² C)1.5mR² D)2mR² E)2mR²/3

Question

Quizplus · Accepted Answer

The moment of inertia of the disk about an axis perpendicular to the disk through the pivot point is found using the parallel axis theorem, which states that $I = I_{cm} + md^2$, where $I_{cm}$ is the moment of inertia about the center of mass, $m$ is the mass, and $d$ is the distance from the center of mass to the new axis. Given $I_{cm} = \frac{1}{2}mR^2$ and $d = R$, the calculation is $I = \frac{1}{2}mR^2 + mR^2 = 1.5mR^2$.

A Uniform Disk (Io = ½ MR2)of Mass M and Radius

A Uniform Disk (I_o = ½ MR²)of Mass M and Radius