The mass per unit area of a circular disc of | vedclass

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(A) Consider a thin circular ring of radius $r$ and thickness $dr$. The area of this ring is $dA = 2 \pi r dr$.The mass of this elemental ring is $dm

Vedclass · Accepted Answer

$2 \pi a^{4} \left( \frac{A}{4} + \frac{aB}{5} \right)$

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(A) Consider a thin circular ring of radius $r$ and thickness $dr$. The area of this ring is $dA = 2 \pi r dr$.The mass of this elemental ring is $dm = \sigma(r) dA = (A + Br) (2 \pi r dr)$.The moment of inertia of this ring about the axis perpendicular to the plane and passing through the centre is $dI = dm r^{2}$.Substituting the value of $dm$,we get $dI = (A + Br) (2 \pi r dr) r^{2} = 2 \pi (A r^{3} + B r^{4}) dr$.To find the total moment of inertia $I$,we integrate $dI$ from $r = 0$ to $r = a$:$I = \int_{0}^{a} 2 \pi (A r^{3} + B r^{4}) dr = 2 \pi \left[ \frac{A r^{4}}{4} + \frac{B r^{5}}{5} \right]_{0}^{a}$.$I = 2 \pi \left( \frac{A a^{4}}{4} + \frac{B a^{5}}{5} \right) = 2 \pi a^{4} \left( \frac{A}{4} + \frac{aB}{5} \right)$.

The mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $\sigma(r) = A + Br$. The moment of inertia of the disc about the axis perpendicular to the plane and passing through its centre is

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