(C) The moment of inertia $I$ of a system of point masses is given by the formula $I = \sum m_i r_i^2$.Here,there are $5$ masses,each of mass $m = 2\,

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(C) The moment of inertia $I$ of a system of point masses is given by the formula $I = \sum m_i r_i^2$.Here,there are $5$ masses,each of mass $m = 2\, kg = 2000\, g$.Each mass is at a distance $r = 10\, cm$ from the axis of rotation.Since all masses are at the same distance,the total moment of inertia is $I = 5 \times m \times r^2$.Substituting the values: $I = 5 \times 2000\, g \times (10\, cm)^2$.$I = 10000 \times 100 = 10^6\, g\cdot cm^2$.

Class 11 Physics System of Particles and Rotational Motion Moment of Inertia and Radius of gyration

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Five masses,each of $2\, kg$,are placed on a horizontal circular disc,which can be rotated about a vertical axis passing through its centre. All the masses are equidistant from the axis at a distance of $10\, cm$. The moment of inertia of the whole system (in $g\cdot cm^2$) is: (Assume the disc is of negligible mass)

A
$10^5$
B
$10^4$
C
$10^6$
D
$10^8$

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System of Particles and Rotational Motion

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Moment of Inertia and Radius of gyration

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Two circular discs $A$ and $B$ are of equal masses and thickness but made of metals with densities ${d_A}$ and ${d_B}$ $({d_A} > {d_B})$. If their moments of inertia about an axis passing through their centres and normal to the circular faces are ${I_A}$ and ${I_B}$ respectively,then:

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