Moment of Inertia of Ring with Sector Removed – Rankers Physics

Moment of Inertia of Ring with Sector Removed

From a circular ring of mass 'M' and radius 'R' an arc corresponding to a $90^\circ$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times $MR^2$. Then the value of 'K' is: (2021)

$\frac{7}{8}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{3}{4}$

Solution:

Since mass is uniformly distributed, removing a $90^\circ$ sector removes $1/4$ of the mass, leaving $3/4$ of the mass at distance $R$. Thus $I = \frac{3}{4}MR^2$, making $K = \frac{3}{4}$.

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