Question

The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis perpendicular to the rod, through its centre is $I$. The moment of inertia of the rod about an axis perpendicular to the rod through its end point is:

• A. $\frac{I}{4}$
• B. $\frac{I}{2}$
• C. $2I$
• D. $4I$

Answer 1

The correct option is D $4I$

We know, a thin uniform rod lying on the axis that is passing through its center have a moment of inertia,

$I_{XY}=\frac{m{L}^2}{12}$

Where,

$m$ is the mass per unit length of the rod and $L$ is the length of the rod.

Let us assume that the endpoint of the rod lies on the point $\left(P,Q\right)$ and the moment of inertia at that point is $I^{\prime }$.

$I=\frac{M{L}^2}{12}$

$M$ is the total mass.

By applying the parallel axis theorem, we have-

$I^{\prime }=I+M{\left(\frac{L}{2}\right)}^2$

Since the distance from the center to the endpoint is $\frac{L}{2}$.

$I^{\prime }=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}$

$=M{L}^2\left(\frac{1}{12}+\frac{1}{4}\right)$

$=M{L}^2\left(\frac{4}{12}\right)$

$=4.\left(\frac{M{L}^2}{12}\right)$

$=4I$