Answer
$I = \frac{1}{2} M R^2$.
Explanation
For a thin hoop/ring, $I = M R^2$; for a uniform solid sphere, $I = \frac{2}{5} M R^2$; for a thin spherical shell, $I = \frac{2}{3} M R^2$.
Home / Library / JEE Main Master Class: Calculus, Mechanics, Coordinate Geometry / Synthetic flashcard
What is the moment of inertia of a uniform solid cylinder/disk of mass $M$ and radius $R$ about its central symmetry axis?
Study flashcard from the deck JEE Main Master Class: Calculus, Mechanics, Coordinate Geometry.