A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of D1, as shown in the figure, will be :

Question

A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of D1, as shown in the figure, will be : (A) 3 MR2 (B) (2/3) MR2 (C) MR2 (D) (4/5) MR2

Tardigrade Admin · Accepted Answer

I = (MR2/2) + 2((MR2/4) + MR2) = (MR2/2) + (MR2/2) + 2 MR2 = 3 MR2

$3 M R^{2}$

$\frac{2}{3} M R^{2}$

$M R^{2}$

$\frac{4}{5} M R^{2}$

$I = \frac{M R ^{2}}{2} + 2 (\frac{M R ^{2}}{4} + M R^{2})$
$= \frac{M R ^{2}}{2} + \frac{M R ^{2}}{2} + 2 M R^{2}$
$= 3 M R^{2}$

3MR2

32​MR2

MR2

54​MR2

I=2MR2​+2(4MR2​+MR2) =2MR2​+2MR2​+2MR2 =3MR2

$3 M R^{2}$

$\frac{2}{3} M R^{2}$

$M R^{2}$

$\frac{4}{5} M R^{2}$

$I = \frac{M R ^{2}}{2} + 2 (\frac{M R ^{2}}{4} + M R^{2})$
$= \frac{M R ^{2}}{2} + \frac{M R ^{2}}{2} + 2 M R^{2}$
$= 3 M R^{2}$