Question

Derive an expression for moment of inertia of a DISC along an axis passing through its centre?
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Answer 1

Answer:

Consider a disc of mass M and radius R. This disc is made up of many infinitesimally small rings as shown in figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inertia (dl) of this small ring is,

dl = (dm) R2

As the mass is uniformly distributed, the mass per unit area (σ) is σ = massareamassarea = MπR2MπR2

The mass of the infinitesimally small ring is,

dm = σ 2πr dr = MπR2MπR2 2πr dr

where, the term (2πr dr) is the area of this elemental ring (2πr is the length and dr is the thickness),

dm = 2MR22MR2 r3 dr

The moment of inertia (I) of the entire disc is,

in the image

hope this answer is helpful for you.

Answer 2

Answer:

MR^2/2

Explanation:

Hope this will help you.