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 place 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 place is :
Updated On: Jul 12, 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 particles
Since, the mass of the disc is negligible therefore, MI of the system = MI of particles
$=5 \times 2 \times(0.1)^{2}=0.1\, kg\, m ^{2}$
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Concepts Used:
System of Particles and Rotational Motion
- The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
- The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
- The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.