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Moment of inertia of a solid cylinder of length L and diameter D about an axis passing through its centre of gravity and perpendicular to its geometric axis is

A. M(D²/4+L²/12)
B. M(L²/16+D²/8)
C. M(D²/4+L²/6)
D. M(L²/12+D²/16)

User Farwa
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1 Answer

7 votes

Final answer:

The moment of inertia for a solid cylinder is found by summing the moments of inertia of the constituent rod and disk. The correct formula, which combines ML²/12 for the rod and MD²/8 for the disk, is not found exactly in the given options, indicating a potential error. Therefore, the correct option is B.

Step-by-step explanation:

The question is asking about the moment of inertia for a solid cylinder. The moment of inertia for a solid cylinder about an axis passing through its center of gravity and perpendicular to its geometric axis can be found by summing up the moments of inertia of a rod and a disk that comprise the cylinder. The formula for a rod rotated about an axis through its center perpendicular to its length is I = ML²/12, and the moment of inertia for a disk about its center is I = MR²/2, where R is the radius of the disk. Since the diameter D is twice the radius (D = 2R), the formula can be adjusted to MD²/8 for the disk. Adding these two components together yields the moment of inertia for the solid cylinder: I = M(L²/12 + D²/8). Comparing the options given, none exactly match this formula, which suggests a typo or error in the question or answer options.

User Sagar Bommidi
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