The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L…

Question

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius **$$R$$** and length **$$L$$** ($$R < L$$) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as **$$M$$**) :

Accepted Answer

C. $$\left(\frac{3}{8}\right)MR^2 + \left(\frac{1}{6}\right)ML^2$$

Answer

A. $$\left(\frac{3}{4}\right)MR^2 + \left(\frac{1}{6}\right)ML^2$$

Answer

B. $$\left(\frac{3}{8}\right)MR^2 + \left(\frac{7}{12}\right)ML^2$$

Answer

D. $$\left(\frac{3}{4}\right)MR^2 + \left(\frac{7}{12}\right)ML^2$$

Answer & step-by-step solution

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