Rotational Motion PYQ — Find x in Moment of Inertia of a Cut Disc 🔥
❓ Question FOR: Let M M and R R be the mass and radius of a (uniform) solid disc. A small disc of radius R / 3 R/3 is removed from the bigger disc (hole cut out) as shown in the figure. The moment of inertia of the remaining part about an axis A B AB passing through the centre O O and perpendicular to the plane of the disc is given as I = x 4 M R 2 . Find the value of x x . (We assume the small disc is concentric with the big disc — i.e. the hole is at the center — this is the standard interpretation unless the figure indicates an off-centre cut.) 🖼️ Question Image ✍️ Short Solution Step 1 — Moment of inertia of the original (full) disc For a uniform solid disc of mass M M and radius R R , I full = 1 2 M R 2 . Step 2 — Mass and moment of inertia of the removed small disc Area (hence mass) scales with R 2 R^2 . The small disc has radius r = R 3 r=\dfrac{R}{3} , so its area is ( r / R ) 2 = ( 1 / 3 ) 2 = 1 / 9 (r/R)^2=(1/3)^2=1/9 ...
