Question

Calculate the moment of inertia of a thin uniform rod of the mass $100\mathrm{g}$and length$60\mathrm{cm}$ about an axis perpendicular to its length and passing through

(a) its center

(b) one end.

Answer 1

Step 1: Given data

Mass of the rod, $\mathrm{M}=100\mathrm{g}=0.1\mathrm{kg}$

Length = $\mathrm{L}=60\mathrm{cm}=0.6\mathrm{m}$

Step 2: Formula used

The moment of inertia about a transverse axis passing through the center of the rod,

$\mathrm{I}=\frac{\mathrm{M}\times {\left(\mathrm{L}\right)}^2}{12}{\mathrm{kgm}}^2$

The moment of inertia about a transverse axis passing through the end of the rod,

$\mathrm{I}=\frac{\mathrm{M}\times {\left(\mathrm{L}\right)}^2}{3}{\mathrm{kgm}}^2$

Where, $\mathrm{I}$ is the moment of inertia, $\mathrm{L}$ is the length of the rod, and $\mathrm{M}$ is the mass of the rod.

Step 3: Calculation of Moment of inertia

The thin uniform rod's moment of inertia about a transverse axis passing through its center is given by

$\mathrm{I}=\frac{\mathrm{M}\times {\left(\mathrm{L}\right)}^2}{12}$

$=\frac{0.1\times {\left(0.6\right)}^2}{12}$

$=$$0.003{\mathrm{kgm}}^2$

The thin uniform rod's moment of inertia about a transverse axis passing through its end is given by,

$\mathrm{I}=\frac{\mathrm{M}\times {\left(\mathrm{L}\right)}^2}{3}$

$=\frac{0.1\times {\left(0.6\right)}^2}{3}$

$=0.012{\mathrm{kgm}}^2$

Therefore, the moment of inertia of center is $0.003{\mathrm{kgm}}^2$ and at the end is $0.012{\mathrm{kgm}}^2$.