Question

The moment of inertia of thin rod is $2\times {10}^{-3}kg{m}^2$ when it is made to rotate about an axis perpendicular to the length of rod and passing through one end. If half length of the rod is cut and removed then the moment of inertia of remaining rod is:

• A. ${10}^{-3}kg{m}^2$
• B. $0.5\times {10}^{-3}kg{m}^2$
• C. $0.25\times {10}^{-3}kg{m}^2$
• D. $25\times {10}^{-3}kg{m}^2$

Answer 1

Answer: (C)

$0.25\times {10}^{-3}kg{m}^2$

Moment of Inertia about one end $=\frac{M{L}^2}{3}$
When half the rod is cut $M^{\prime }=\frac{M}{2}$ and $L^{\prime }=\frac{L}{2}$
$\therefore I=\frac{1}{3}\frac{M}{2}{\left(\frac{L}{2}\right)}^2=\frac{M{L}^2}{3\left(8\right)}=\frac{I}{8}=\frac{2\times {10}^{-3}}{8}kg{m}^2$
$=0.25\times {10}^{-3}kg{m}^2$