Question

Five particles of mass 2 kg are attached to the rim of a circular disc of radius 0.1 m and negligible mass. Moment of inertia of the system about the axis passing through the center of the disc and perpendicular to its plane is.

• A. $1kg{m}^2$
• B. $0.1kg{m}^2$
• C. $2kg{m}^2$
• D. $0.2kg{m}^2$

Answer 1

The correct option is B $0.1kg{m}^2$

We will not consider the moment of inertia of ring because it doesn't have any mass. So, moment of inertia of five particle system $I=5m{r}^2=5\times 2\times (0.1{)}^2=0.1kg-m^2.$
Note: The masses are concentrated at fixed distance from the axis similar to that of ring. That is why you simply need to add all the individual contributions.