Question:
Five particles of mass $2\,kg$ are attached to the rim of a circular disc of radius $0.1\,m$ and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is:
Five particles of mass $2\,kg$ are attached to the rim of a circular disc of radius $0.1\,m$ and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is:
Updated On: Jun 7, 2022
- $ 1\,kg\,m^{2} $
- $ 0.1\,kg\,m^{2} $
- $ 2\,kg\,m^{2} $
- $ 0.2\,kg\,m^{2}$
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The Correct Option is B
Solution and Explanation
The moment of inertia of the given system that contains $5$ particles each of mass $=2\, kg$ on the rim of circular disc of radius $0.1 m$ and of negligible mass is given by
$= MI$ of disc $+ MI$ of particle
Since the mass of the disc is negligible therefore, $MI$ of the system $= MI $ of particle
$=5 \times 2 \times(0.1)^{2}=0.1 \,kg \,m ^{2}$
$= MI$ of disc $+ MI$ of particle
Since the mass of the disc is negligible therefore, $MI$ of the system $= MI $ of particle
$=5 \times 2 \times(0.1)^{2}=0.1 \,kg \,m ^{2}$
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