Practice portal › Rotational Motion › Moment of Inertia
Asked in JEE Main 8th April 1st Shift 2019 · Moment of inertia by integration
Take a ring of radius r and width dr. Its area is 2π r dr and its mass is ρ₀ r· 2π r dr.
Total mass: M=∫₀^R 2πρ₀ r²dr=(2πρ₀R³)/3.
Moment of inertia about the central perpendicular axis: I_c=∫₀^R r²·2πρ₀r²dr=(2πρ₀R⁵)/5.
Divide to remove ρ₀: (I_c)/M=(R⁵/5)/(R³/3)=(3R²)/5, so I_c=3/5MR².
Now shift to the edge, a distance R away: I=I_c+MR²=3/5MR²+MR²=8/5MR².
a=8/5.
Sanity check: 3/5>1/2, as it must be — the density rises with r, pushing mass outwards compared with a uniform disc.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer