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Question

Two uniform identical rods each of mass M and length L are joined to form a cross as shown in the figure. Find the moment of inertia of the cross about a bisector in the plane of rods as shown by dotted line in the figure.

A
ML212
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B
ML24
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C
ML26
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D
ML23
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Solution

The correct option is A ML212
Given, length of the rod is L and mass of the rod is M

As we know that, Iz of a rod is given as ML212
So, MOI of the two rod shown will be (2×Iz), which is ML26
Now, from the perpendicular axis theorem we have
Iz=Ix+Iy
and due to symmetry, Ix=Iy
Iz=2Ix or Ix=Iy=Iz2
Hence, MOI of the cross about the bisector will be
(ML26)2=ML212

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