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

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) :

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G1 Physics · Accepted Answer

Each cylindrical side has mass M' = M/4. The required axis lies in the plane of the square and passes through the midpoints of two opposite sides. There are two cylinders whose centres lie on the axis and two cylinders parallel to the axis but displaced by L/2. For each of the two cylinders whose centre lies on the axis, the moment of inertia about the required axis is I₁ = M'R²/4 + M'L²/12. For each of the other two cylinders, the parallel-axis theorem gives I₂ = M'R²/2 + M'(L/2)². Therefore,…

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

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) :