Question

126.Three identical rings each of mass m are placed in such a manner so that their centres are vertices
of an equilateral triangle of edge 2r. The moment of inertia about an axis passing through centre of
triangle and perpendicular to plane is kmrº, where k is
1)3
2)5
3) 7
4) 9​

Answer 1

$\Large\boxed{\sf{\quad(3)\quad\!7}\quad}$

Since an equilateral triangle of side '2r' is formed with the three identical rings, the radius of each ring is 'r'.

As shown in the fig., the center of rings are located at points A(0, 0), B(2r, 0) and C(r, r√3). The center of mass is located at O which is usually the centroid of the triangle since the rings are identical. Then,

$\longrightarrow\sf{O=\left(\dfrac{0+r+2r}{3},\ \dfrac{0+r\sqrt3+0}{3}\right)}$

$\longrightarrow\sf{O=\left(r,\ \dfrac{r}{\sqrt3}\right)}$

Since the triangle is equilateral, $\sf{OA=OB=OC=x}$ and is given by,

$\longrightarrow\sf{x=\sqrt{r^2+\left(\dfrac{r}{\sqrt3}\right)^2}}$

$\longrightarrow\sf{x=\sqrt{r^2+\dfrac{r^2}{3}}}$

$\longrightarrow\sf{x=\sqrt{\dfrac{4r^2}{3}}}$

$\longrightarrow\sf{x=\dfrac{2r}{\sqrt3}}$

We know that the moment of inertia of each ring about an axis passing through their respective centers and perpendicular to their plane, is,

$\longrightarrow\sf{I_r=mr^2}$

But by theorem of parallel axes, its moment of inertia about an axis passing through the center of the triangle, i.e., O, and perpendicular to the plane, is,

$\longrightarrow\sf{I_r'=mr^2+mx^2}$

This is the same for each ring since they're identical and are at same distance away from O.

Hence the net moment of inertia of the system,

$\longrightarrow\sf{3(mr^2+mx^2)=k\,mr^2}$

$\longrightarrow\sf{3m(r^2+x^2)=k\,mr^2}$

$\longrightarrow\sf{3\left(r^2+\dfrac{4r^2}{3}\right)=k\,r^2}$

$\longrightarrow\sf{3\cdot\dfrac{7r^2}{3}=k\,r^2}$

$\longrightarrow\sf{7r^2=k\,r^2}$

Therefore,

$\longrightarrow\large\text{ \sf{\underline{\underline{k=7}}} }$