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
$\Rightarrow I_{1}-I_{3}=M\left((R-y)^{2}-(R+y)^{2}\right)$
$=-4 M R y$ ...(iii)

Similarly, from above figure by parallel axes theorem, we have
$I_{2}=I_{c}+M(R-x)^{2}$
and $I_{4}=I_{c}+M(R+x)^{2}$
$\Rightarrow I_{2}-I_{4}=-4 M R x$ ...(iv)
So, from Eqs. $(iii)$ and $(iv),$ we have
$\left(I_{1}-I_{3}\right)^{2}+\left(I_{2}-I_{4}\right)^{2}=16 M^{2} R^{2}\left(x^{2}+y^{2}\right)$
$\Rightarrow \sqrt{x^{2}+y^{2}}=\frac{1}{4 M R} \sqrt{\left(I_{1}-I_{3}\right)^{2}+\left(I_{2}-I_{4}\right)^{2}}$
But $\sqrt {x^{2}+y^{2}} =$ distance of centre of mass from centre of the disc