JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Rotational Mechanics

A thin disc of mass

M

and radius

R

has mass per unit area

σ (r) = k r^{2}

where

r

is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is:

Answer: (D)

Solution

The disc has a surface density that grows as

σ = k r^{2}

, so we integrate ring by ring. The moment of inertia is

I = \int_{0}^{R} r^{2} d m = \int_{0}^{R} r^{2} (k r^{2}) (2 π r d r) = 2 π k \int_{0}^{R} r^{5} d r = \frac{2 π k R ^{6}}{6}

, while the total mass is

M = \int_{0}^{R} (k r^{2}) (2 π r d r) = 2 π k \int_{0}^{R} r^{3} d r = \frac{2 π k R ^{4}}{4}

. Dividing one by the other,

I = \frac{2}{3} M R^{2}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option