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(B) The moment of inertia $I$ of a system of discrete particles about an axis is given by the sum of the products of the mass of each particle and the

Vedclass · Accepted Answer

$\frac{m{l^2}}{6}n(n + 1)(2n + 1)$

Vedclass · Answer

(B) The moment of inertia $I$ of a system of discrete particles about an axis is given by the sum of the products of the mass of each particle and the square of its perpendicular distance from the axis: $I = \sum m_i r_i^2$.Here,we have $n$ particles each of mass $m$ located at distances $\ell, 2\ell, 3\ell, \dots, n\ell$ from the axis $OA$.Therefore,the total moment of inertia is:$I = m(\ell)^2 + m(2\ell)^2 + m(3\ell)^2 + \dots + m(n\ell)^2$$I = m\ell^2 (1^2 + 2^2 + 3^2 + \dots + n^2)$Using the standard formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$I = m\ell^2 \left[ \frac{n(n+1)(2n+1)}{6} \right]$$I = \frac{m\ell^2}{6} n(n+1)(2n+1)$

Find the moment of inertia of the system of particles about the axis $OA$ passing through point $O$.

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