    Question

# Four point masses, each of mass m are present at the four corners of a square of side length equal to a. Axis of rotation is passing through through one particle as shown in the figure below. The moment of inertia about the axis of rotation is A
4 ma2
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B
3 ma2
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C
(2+2) ma2
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D
22 ma2
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Solution

## The correct option is D 3 ma2As we can see from the figure, axis of rotation passes through the point mass 4. Hence MOI for point mass 4 will be zero. Distance of point mass 1 and 3 from the axis of rotation =acos45∘=a√2 Distance of point mass 2 from the axis of rotation is =√a2+a2=√2a Hence, moment of inertia I=I1+I2+I3+I4 I=m(a√2)2+m(√2a)2+m(a√2)2+0 I=ma22+2ma2+ma22+0 I=3ma2  Suggest Corrections  1      Similar questions  Explore more