Question

Moment of inertia of a thin rod of mass $M$ and length about an axis passing through its centre is $\frac{M L^{2}}{12}$. Its moment of inertia about a parallel axis at a distance of $\frac{L}{4}$ from this axis is given by

• A. $\frac{M L^{2}}{48}$
• B. $\frac{M L^{3}}{48}$
• C. $\frac{M L^{2}}{12}$
• D. $\frac{7 M L^{2}}{48}$

Answer 1

Answer: (D)

Apply parallel axis theorem,
$I=I_{C M}+M h^{2}$, we get
$\Rightarrow =\frac{M L^{2}}{12}+M\left(\frac{L}{4}\right)^{2}$
$=\frac{7 M L^{2}}{48}$