Four similar point masses ( m each) are symmetrically placed on the circumference of a disc of mass M and radius R . Moment of inertia of the system about an axis passing through centre O and perpendicular to the plane of the disc will be

Question

Four similar point masses ( m each) are symmetrically placed on the circumference of a disc of mass M and radius R . Moment of inertia of the system about an axis passing through centre O and perpendicular to the plane of the disc will be (A) MR2+4mR2 (B) MR2+(8/5)mR2 (C) mR2+8MR2 (D) (M R2/2)+4mR2

Tardigrade Admin · Accepted Answer

Moment of inertia of the disc passing through the centre perpendicular to the plane is given by,

I_{D} = \frac{M R ^{2}}{2}

.
Each point mass is separated by a distance

R

from the axis, so moment of inertia of all the point masses is,

I_{p} = 4 m R^{2}

Total moment of inertia of the system about the axis is,

I = I_{D} + I_{p}

\Rightarrow I = \frac{M R ^{2}}{2} + 4 m R^{2}

Q. Four similar point masses ( $m$ each) are symmetrically placed on the circumference of a disc of mass $M$ and radius $R$ . Moment of inertia of the system about an axis passing through centre $O$ and perpendicular to the plane of the disc will be

$M R^{2} + 4 m R^{2}$

$M R^{2} + \frac{8}{5} m R^{2}$

$m R^{2} + 8 M R^{2}$

$\frac{M R ^{2}}{2} + 4 m R^{2}$

Q. Four similar point masses ( m each) are symmetrically placed on the circumference of a disc of mass M and radius R . Moment of inertia of the system about an axis passing through centre O and perpendicular to the plane of the disc will be

MR2+4mR2

MR2+58​mR2

mR2+8MR2

2MR2​+4mR2

Q. Four similar point masses ( $m$ each) are symmetrically placed on the circumference of a disc of mass $M$ and radius $R$ . Moment of inertia of the system about an axis passing through centre $O$ and perpendicular to the plane of the disc will be

$M R^{2} + 4 m R^{2}$

$M R^{2} + \frac{8}{5} m R^{2}$

$m R^{2} + 8 M R^{2}$

$\frac{M R ^{2}}{2} + 4 m R^{2}$