Question

Two point size bodies of masses $20g$, $30g$ are fixed at two ends of a light rod of length $1.5m$. The moment of inertia of these two bodies about an axis perpendicular to the length of rod and passing through their center of mass is:

• A. $1.8\times {10}^{-2}kg{m}^2$
• B. $2.7\times {10}^{-2}kg{m}^2$
• C. $4.5\times {10}^{-2}kg{m}^2$
• D. $6\times {10}^{-2}kg{m}^2$

Answer 1

Answer: (B)

$2.7\times {10}^{-2}kg{m}^2$

The position of COM from $20g$ mass is $30\times 1.5/(20+30)=0.9m$

So MI of $20g$ mass about COM is $20\times {10}^{-3}\times {0.9}^2=1.62\times {10}^{-2}kg{m}^2$

and MI of $30g$ mass about COM is $30\times {10}^{-3}\times (1.5-0.9{)}^2=1.08\times {10}^{-2}kg{m}^2$

So total MI is $(1.62+1.08)\times {10}^{-2}=2.7\times {10}^{-2}kg{m}^2$