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):
- 34MR2 + 16ML2
- 38MR2 + 712ML2
- 34MR2 + 712ML2
- 38MR2 + 16ML2
Answer
(4) 38MR2 + 16ML2
Solution
Mass of each cylinder = M4. The axis lies in the plane of the loop and passes through the mid points of two opposite sides.
Two sides are cut by the axis at their centres, perpendicular to their own length:
I1 = 2 M4 ( R24 + L212 ) = MR28 + ML224
The other two sides are parallel to the axis at a distance L2 (parallel axis theorem):
I2 = 2 M4 ( R22 + L24 ) = MR24 + ML28
I = I1 + I2 = ( 18 + 14 )MR2 + ( 124 + 324 )ML2
I = 38MR2 + 16ML2
Answer: (4)