Question

Five particles of masses $2kg$ each are attached to the rim of a circular disc of radius $0.1m$ and negligible mass. Moment of inertia of the system about the axis passing through the centre 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 $0.1kg-m^2$

Given,

Five particles, all have equal mass $M=2kg$

All have equal mass and equal distance from axis $R=0.1m$

Then, Moment of each particle will be equal to $I=M{R}^2=2\times {0.1}^2=0.02kg-m^2$

Total moment of inertia of all the five particles $I_{total}=5\times I=5\times 0.02=0.1kg-m^2$

Total moment of inertia of all the five particles$I_{total}=0.1kg-m^2$