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the moment of inertia of semicircular ring about an axis which is perpendicular to the plane of the ring and passes through centre
Asked by raghuwanshidaksh9 | 14 Oct, 2019, 05:48: PM Expert Answer Figure shows a semi-circular ring  of radius R.

Let the centre of curvature O coincides with origin of polar coordinate (r-θ) system .

It is required to get moment of inertia about axis passing through P which is the geometrical centre of semi-circular ring.
Axis of rotation is perpendicular to the plane of ring.

Let us consider a small length dl of ring at Q .  If the line joining Q to origin makes angle θ with reference line
and length dl subtends angle dθ, then this small length equals ( R dθ).

If M is mass of whole ring and mass is uniformly distributed, mass density m, i.e. mass per unit length is given by , m = M/(πR)

If Polar coordinates of point Q is (Rcosθ , Rsinθ) and that of point P (0,R) then distance PQ is given by, Moment of inertia of small length at Q,  dI  = dm r2 .............(1)

where dm is mass of small length,  m×dl = m×R dθ ...................(2)

Hence moment of inertia I of whole ring is given by, I = 0.363 M R2

Answered by Thiyagarajan K | 15 Oct, 2019, 09:22: AM
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