Question

Four rings each of mass M and radius R are arranged
as shown in the figure. The moment of inertia of the
system about the axis yy' is
(2) 3MR2
(1) 2MR2
(3) 4MR2
(4) 5MR2​

Answer 1

The moment of inertia of a ring of radius R and mass M about an axis passing through its diameter is $\sf{\dfrac{1}{2}\,MR^2.}$

Consider the rings at top and bottom of the system. The axis is passing through the diameter so the moment of inertia is $\sf{I_1=\dfrac{1}{2}\,MR^2}$ for each.

Consider the two rings in the middle of the system. The axis is tangential to them parallel to its diameter, so by parallel axes theorem the moment of inertia for each of the two rings is,

$\sf{\mapsto I_2=\dfrac{1}{2}\,MR^2+MR^2}$

$\sf{\mapsto I_2=\dfrac{3}{2}\,MR^2}$

Hence the moment of inertia of the whole system is,

$\sf{\longrightarrow I=2[I_1+I_2]}$

$\sf{\longrightarrow I=2\left[\dfrac{1}{2}\,MR^2+\dfrac{3}{2}\,MR^2\right]}$

$\sf{\longrightarrow\underline{\underline{I=4MR^2}}}$

Answer 2

Solution :-

please refer the attachment first

◕ For ring 1 and 4 ,

The moment of inertia of the ring about any diameter at the axis is given by ,

➜ I = MR² / 2

Hence ,

➜ I = I 1 = I 4 = MR² / 2

◕ For ring 2 and 3

Now by applying parallel axes theorem ,

The moment of inertia of the disc about an axis with the tangent in it's plane is ,

➜ I ' = I + MR²

➜ I' = MR² / 2 + MR²

➜ I' = 3 MR² /2

➜ I' = I 2 = I 3 = 3 MR² / 2

The total moment of inertia of the system about the yy' axis

➜ I * = I 1 + I 2 + I 3 + I 4

➜ I * = MR² /2+ 3MR² / 2 + 3MR² / 2 + MR²/2

➜ I * = MR² + 3MR²

➜ I * = 4MR²

The moment of inertia of the  system about the axis yy' is 4MR²