Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. The moment of inertia of this system about an axis passing through one of the particle and perpendicular to plane of disc is

Question

Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. The moment of inertia of this system about an axis passing through one of the particle and perpendicular to plane of disc is (A) 16mR2 (B) (3M+16m) (R2/2) (C) (3m+12M) (R2/2) (D) zero

Tardigrade Admin · Accepted Answer

According to the theorem of parallel axes, moment of inertia of disc about an axis passing through K and perpendicular to plane of disc is

= \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2}

Total moment of inertia of the system

= \frac{3}{2} M R^{2} + m (2 R)^{2} + m (2 R)^{2} + m (2 R)^{2}

= (3 M + 16 m) \frac{R ^{2}}{2}

Q. Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. The moment of inertia of this system about an axis passing through one of the particle and perpendicular to plane of disc is

$16 m R^{2}$

$(3 M + 16 m) \frac{R ^{2}}{2}$

$(3 m + 12 M) \frac{R ^{2}}{2}$

zero

Q. Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. The moment of inertia of this system about an axis passing through one of the particle and perpendicular to plane of disc is

16mR2

(3M+16m)2R2​

(3m+12M)2R2​

zero

According to the theorem of parallel axes, moment of inertia of disc about an axis passing through K and perpendicular to plane of disc is =21​MR2+MR2=23​MR2 Total moment of inertia of the system =23​MR2+m(2R)2+m(2​R)2+m(2​R)2 =(3M+16m)2R2​

$16 m R^{2}$

$(3 M + 16 m) \frac{R ^{2}}{2}$

$(3 m + 12 M) \frac{R ^{2}}{2}$