Question

Four identical particles of mass 0.50kg each are placed at the vertices of a$\mathbf{2}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{0}\mathit{m}$ square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

Short answer

1. The rotational inertia of arigid body about an axis that passes through the midpoint of opposite sides and lies in the plane of the square is 2.0kg.m² .
2. The rotational inertia of a rigid body about an axis that passes through midpoint of one of the sides and is perpendicular to the plane of the square is 6.0kg.m² .
3. What is the rotational inertia of a rigid body about an axis that lies in the plane of the square and passes through two diagonally opposite particles is 2.0kg.m².

Step-by-step solution

Understanding the given information

1. Mass of each particle is, m = 0.50kg.
2. Distance between two masses is 2m.

Concept and formula used in the given question

Rotational inertia is a constant for a particular rigid body and a given axis of rotation. By using the concept of the rotational moment of inertia, you can calculate the moment of inertia when the axis is passing through various points.

$\mathit{l}\mathbf{=}\mathbf{\sum }{\mathit{m}}_{\mathbf{i}}{\mathit{r}}_{\mathbf{i}}^{\mathbf{2}}$

(a) Calculation for therotational inertia of this rigid body about an axis that passes through the midpoints of opposite sides and lies in the plane of the square

Each mass is placed at r = 1m from axis so, you can calculate rotational inertia by the formula

$$\begin{gathered} l=\sum m_i{r}_i^2 \\ =0.50kg\times 1m^2+0.50kg\times 1m^2+0.50kg\times 1m^2+0.50kg\times 1m^2 \\ =2.0kg.m^2 \end{gathered}$$

Hence the theoretical inertia of rigid body is, 2.0kg.m².

(b) Calculation for therotational inertia of this rigid body about an axis thatpasses through the midpoint of one of the sides and is perpendicular to the plane of the square

When the axis passes through mid-point of one of the sides of square, in this case, two masses nearest the axis are r = 1m away from it, but two are at distance

$r=\sqrt{{\left(1m\right)}^2+{\left(2m\right)}^2}i.e.,r=\sqrt{5}mfromit.$

So rotational moment of inertia

$\begin{array}{rcl}l& =& \sum m_i{r}_i^2\\ & =& \left(0.50kg\times 1m^2\right)+\left(0.50kg\times 1m^2\right)+\left(0.50kg\times 1m^2\right)+\left(0.50kg\times 1m^2\right)\\ & =& 6.0kg.m^2\end{array}$

Hence the theoretical inertia of rigid body is, 6.0kg.m².

(c) Calculation for therotational inertia of this rigid body about an axis that lies in the plane of the square and passes through two diagonally opposite particles

When the axis lies in plane of the square and passes through two diagonally opposite particles so, r = 0 for this particle. And $r=\sqrt{{\left(1m\right)}^2+{\left(1m\right)}^2}=\sqrt{2}$for another of the two particles.

So rotational moment of inertia,

$$\begin{gathered} l=\sum m_i{r}_i^2 \\ =0.50kg\times 0+0.50kg\times 0+0.50kg\times 2m^2+0.50kg\times 2m^2 \\ =2.0kg.m^2 \end{gathered}$$

Hence the theoretical inertia of rigid body is, 2.0kg.m² .