Question

Calculate the rotational inertia of a meter stick, with mass 0.56kg, about an axis perpendicular to the stick and located at the 20cmmark. (Treat the stick as a thin rod.)

Short answer

The rotational inertia of meter stick is, 0.097kg.m².

Step-by-step solution

Understanding the given information

The mass of the meter stick is, m = 0.56kg

The axis is located at 0.2m .

Concept and Formula used

By finding the moment of inertia for the meter stick about its center of mass and then applying the parallel axis theorem to it, we can find moment of inertia about an axis located at 20 cm mark.

1. Parallel axis theorem$\mathit{l}\mathbf{=}{\mathit{l}}_{\mathbf{c}\mathbf{o}\mathbf{m}}\mathbf{+}\mathit{m}{\mathit{h}}^{\mathbf{2}}$
2. Moment of inertia of a thin rod about an axis at passing through the center of mass${\mathit{l}}_{\mathbf{c}\mathbf{o}\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{12}}\mathit{m}{\mathit{L}}^{\mathbf{2}}$

Calculation for the rotational inertia of a meter stick

We are given a meter sick; we need to find moment of inertia (I) about an axis as shown in the figure below:

In the figure there are two axes, one which is passing through center of mass $l_{com}$,and as it is a meter stick, it is located at 0.5m.

The other axis at which we need to find the moment of inertia (I) is located at 0.2m.

Distance between the two axes h = 0.5m - 0.2 = 0.3m

Let us find the moment of inertia at center of mass l_{com ,}as we are considering the meter stick as a thin rod

$\begin{array}{rcl}{l}_{com}& =& \frac{1}{12}m{L}^2\\ & =& \frac{1}{12}\times 0.56kg\times {\left(1m\right)}^2\\ l_{com}& =& 0.0466kh.m^2\end{array}$ …(1)

Now by applying Parallel axis theorem,

$l=l_{com}+m{h}^2$

Substitute all the value in the above equation.

$\begin{array}{rcl}l& =& 0.0466kg.m^2+0.56kg\times {\left(0.3m\right)}^2\\ & =& 0.097kg.m^2\end{array}$

Hence the moment of inertia of the meter stick about an axis located at 0.2m is 0.097kg.m² .