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Q. Out of the given bodies (of same mass) for which the moment of inertia will be maximum about the axis passing through its centre of gravity and perpendicular to its plane?

Rajasthan PMTRajasthan PMT 2005System of Particles and Rotational Motion

Solution:

For disc, $I=\frac{1}{2} m a^{2}$
For ring, $I=m a^{2}$
For square of side $2 a$
$=\frac{M}{12}\left[(2 a)^{2}+(2 a)^{2}\right]=\frac{2}{3} M a^{2}$
For square of rod of length $2 a$
$I=4\left[M \frac{(2 a)^{2}}{12}+M a^{2}\right]=\frac{16}{3} M a^{2}$
Hence, moment of inertia is maximum for square of four rods.