Question

The moments of inertia of a non-uniform circular disc (of mass $M$ and radius $R$) about four mutually perpendicular tangents $AB,BC,CD,DA$ are $I_1,I_2,I_3$ and $I_4,$ respectively (the square $ABCD$ circumscribes the circle). The distance of the centre of mass of the disc from its geometrical centre is given by

• A. $\frac{1}{4MR}\sqrt{{\left(I_1-I_3\right)}^2+{\left(I_2-I_4\right)}^2}$
• B. $\frac{1}{12MR}\sqrt{{\left(I_1-I_3\right)}^2+{\left(I_2-I_4\right)}^2}$
• C. $\frac{1}{3MR}\sqrt{{\left(I_1-I_2\right)}^2+{\left(I_3-I_4\right)}^2}$
• D. $\frac{1}{2MR}\sqrt{{\left(I_1-I_3\right)}^2+{\left(I_2+I_4\right)}^2}$

Answer 1

Answer: (A)

Let $O$ is centre of disc and $C$ is its centre of mass.

If $I_c$ = moment of inertia about an axis through centre of mass of disc, $I_1$ = moment of inertia about an tangential axis $AB,$
$I_3$ = moment of inertia about an tangential axis $CD,$
$M$= mass of disc and $R=$ radius of disc.
Then, by parallel axes theorem, we have
$I_1=I_c+M\cdot (R-y{)}^2$ ...(i)
$I_3=I_c+M\cdot (R+y{)}^2$ ...(ii)
Subtracting Eqs. $(i)$ and $(ii),$ we get
$\Rightarrow I_1-I_3=M\left((R-y{)}^2-(R+y{)}^2\right)$
$=-4MRy$ ...(iii)

Similarly, from above figure by parallel axes theorem, we have
$I_2=I_c+M(R-x{)}^2$
and $I_4=I_c+M(R+x{)}^2$
$\Rightarrow I_2-I_4=-4MRx$ ...(iv)
So, from Eqs. $(iii)$ and $(iv),$ we have
${\left(I_1-I_3\right)}^2+{\left(I_2-I_4\right)}^2=16M^2{R}^2\left(x^2+y^2\right)$
$\Rightarrow \sqrt{x^2+y^2}=\frac{1}{4MR}\sqrt{{\left(I_1-I_3\right)}^2+{\left(I_2-I_4\right)}^2}$
But $\sqrt{x^2+y^2}=$ distance of centre of mass from centre of the disc