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Two uniform, thin identical rods each of mass M and length L are joined at middle so as to form a cross as shown. The moment of inertia of the cross about a bisector line EF (in the plane of cross) is
detailed solution
Correct option is C
We have to find I1? I1 = I2----------(1) I0 = ML212+ML212 = ML26now I1 +I2 = I0--------(2)From (1) and (2), I1 = I02 = ML212Talk to our academic expert!
Similar Questions
The moment of inertia of a circular disc of radius 2 m and mass 2 kg, about an axis passing through its centre of mass is 2 kg – . Its moment of inertia about an axis parallel to this axis and passing through its edge (in Kg -) is
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