Moment of inertia of an annular disc of mass M having inner and outer radius r and R respectively about an axis passing through centre of mass and perpendicular to the plane of the disc is

Question

Moment of inertia of an annular disc of mass M having inner and outer radius r and R respectively about an axis passing through centre of mass and perpendicular to the plane of the disc is (A)  (1/2)M(R2+r2) (B)  (1/2)M(R2-r2) (C) M(R2+r2) (D) M(R2-r2)

Tardigrade Admin · Accepted Answer

I = ∫Rr(M/π(R2-r2))(2π xdx)x2 =(M/2(R2-r2))(R4-r4) = (M/2)(R2+r2)

Q. Moment of inertia of an annular disc of mass $M$ having inner and outer radius $r$ and $R$ respectively about an axis passing through centre of mass and perpendicular to the plane of the disc is

$\frac{1}{2} M (R^{2} + r^{2})$

$\frac{1}{2} M (R^{2} - r^{2})$

$M (R^{2} + r^{2})$

$M (R^{2} - r^{2})$

$I = \int_{r}^{R} \frac{M}{π ( R ^{2} - r ^{2} )} (2 π x d x) x^{2}$
$= \frac{M}{2 ( R ^{2} - r ^{2} )} (R^{4} - r^{4})$
$= \frac{M}{2} (R^{2} + r^{2})$

Q. Moment of inertia of an annular disc of mass M having inner and outer radius r and R respectively about an axis passing through centre of mass and perpendicular to the plane of the disc is

21​M(R2+r2)

21​M(R2−r2)

M(R2+r2)

M(R2−r2)