Question

Four particles each of mass M are lying
symmetrically on the rim of a disc of mass 6M
and radius R. Moment of inertia of this system
about an axis passing through one of the particles
and perpendicular to plane of disc is​

Answer 1

Answer:

Explanation:13 mr^2/2

Answer 2

Answer:

The moment of inertia of the system is

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

Explanation:

Moment of inertia of the disc about the axis

$=\frac{MR^2}{2}+MR^2$

$=\frac{3MR^2}{2}$

Moment of inertia of particle 1

$=m\times (R\sqrt{2})^2$

$=2mR^2$

Similarly moment of inertia of particle 1

$=2mR^2$

Moment of inertia of particle 3

$=m(2R)^2$

$=4mR^2$

Moment of inertia of particle 4 = 0

The moment of inertia of the system

$=\frac{3MR^2}{2}+2mR^2+2mR^2+4mR^2+0$

$=\frac{3}{2}MR^2+8mR^2$

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

Hope this helps.