Question:
The moment of inertia of a circular ring of mass $1\,kg $ about an axis passing through its centre and perpendicular to its plane is $4\, kg-m^{2}$ . The diameter of the ring is
The moment of inertia of a circular ring of mass $1\,kg $ about an axis passing through its centre and perpendicular to its plane is $4\, kg-m^{2}$ . The diameter of the ring is
Updated On: May 15, 2024
- 2 m
- 4 m
- 5 m
- 6 m
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The Correct Option is B
Solution and Explanation
Moment of inertia of circular ring about an axis passing through its center of mass and perpendicular to its plane
$I=M R^{2}$
here $I=4\, k g-m^{2}$
$m =1\, kg$
$=R^{2}=\frac{4}{1}=4 R=2\, m$
Therefore, diameter of ring $=4\, m$
$I=M R^{2}$
here $I=4\, k g-m^{2}$
$m =1\, kg$
$=R^{2}=\frac{4}{1}=4 R=2\, m$
Therefore, diameter of ring $=4\, m$
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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.