Question

From a circular ring of mass M and radius R. an are corresponding to a 90° 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. Then the value of k is​

Answer 1

Let the mass of the removed arc be m. So the moment of inertia of the remaining portion should be,

$\quad$

$\mapsto\sf{I=MR^2-mR^2=k\cdot MR^2}$

$\quad$

The length of the arc should be,

$\quad$

$\mapsto\sf{l=R\cdot\dfrac {\pi}{2}}$

$\quad$

since the arc is corresponding to a 90° or a (π/2) radian sector.

$\quad$

Since the ring is assumed to be uniform, the linear density is the same everywhere. Thus,

$\quad$

$\mapsto\sf{\dfrac {M}{2\pi R}=\dfrac{m}{\left (\dfrac {R\pi}{2}\right)}}\\\\\\\mapsto\sf{m=\dfrac {M}{4}}$

$\quad$

Then,

$\quad$

$\mapsto\sf{MR^2-\left (\dfrac {M}{4}\right)R^2}=k\cdot MR^2\\\\\\\mapsto\sf{\dfrac {3}{4}MR^2=k\cdot MR^2}$

$\quad$

Therefore,

$\quad$

$\mapsto\sf{\underline {\underline {k=\dfrac {3}{4}}}}$