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Question

A ring having linear mass density 5 kg/m and radius 0.5 m is rotated about the axis passing through it's centre and perpendicular to it's plane. Find the moment of inertia (MI) of the ring about the required axis.

A
0.5 kg-m2
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B
1 kg-m2
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C
2.5π kg-m2
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D
1.25π kg-m2
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Solution

The correct option is D 1.25π kg-m2
Mass of ring,(m) = linear mass density × circumference
r=0.5 m
m=5×(2πr)=5×(2×π×0.5)
m=5π kg

MI of ring about the required axis YY
as shown in figure is:
I=mr2
I=(5π)(0.5)2
I=1.25π kg-m2
So, moment of inertia of ring about the axis YY is 1.25π kg-m2

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