# 18.6 Moments of inertia of rigid bodies  (Page 2/5)

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Calculation of elemental mass "đm" makes use of appropriate density on the basis of the nature of mass distribution in the rigid body :

$\begin{array}{l}\mathrm{mass}=\phantom{\rule{2pt}{0ex}}\mathrm{appropriate density}\phantom{\rule{4pt}{0ex}}x\phantom{\rule{4pt}{0ex}}\mathrm{geometric dimension}\end{array}$

The choice of density depends on the nature of body under consideration. In case where element is considered along length like in the case of a rod, rectangular plate or ring, linear density (λ) is the appropriate choice. In cases, where surface area is involved like in the case of circular plate, hollow cylinder and hollow sphere, areal density (σ) is the appropriate choice. Finally, volumetric density (ρ) is suitable for solid three dimensional bodies like cylinder and sphere. The elemental mass for different cases are :

$\begin{array}{l}đm=\lambda đx\\ đm=\sigma đA\\ đm=\rho đV\end{array}$

The MI integral is then expressed by suitably replacing "dm" term by density term in the integral expression. This approach to integration using elemental mass assumes that mass distribution is uniform.

## Evaluation of moment of inertia

In this section, we shall determine MI of known geometric bodies about the axis of its symmetry.

## Mi of a uniform rod about its perpendicular bisector

The figure here shows the small element with repect to the axis of rotation. Here, the steps for calculation are :

(i) Infinetesimally small element of the body :

Let us consider an small element "dx" along the length, which is situated at a linear distance "x" from the axis.

(ii) Elemental mass :

Linear density, λ, is the appropriate density type in this case.

$\begin{array}{l}\lambda =\frac{M}{L}\end{array}$

where "M" and "L" are the mass and length of the rod respectively. Elemental mass (đm) is, thus, given as :

$\begin{array}{l}đm=\lambda đx=\left(\frac{M}{L}\right)đx\end{array}$

(iii) Moment of inertia for elemental mass :

Moment of inertia of elemental mass is :

$\begin{array}{l}đI={r}^{2}đm={x}^{2}\left(\frac{M}{L}\right)đx\end{array}$

(iv) Moment of inertia of rigid body :

$\begin{array}{l}I=\int {r}^{2}đm=\int {x}^{2}\left(\frac{M}{L}\right)đx\end{array}$

While setting limits we should cover the total length of the rod. The appropriate limits of integral in this case are -L/2 and L/2. Hence,

$\begin{array}{l}I={\int }_{\frac{-L}{2}}^{\frac{L}{2}}\left(\frac{M}{L}\right){x}^{2}đx\end{array}$

Taking the constants out of the integral sign, we have :

$\begin{array}{l}⇒I=\left(\frac{M}{L}\right){\int }_{\frac{-L}{2}}^{\frac{L}{2}}{x}^{2}đx\end{array}$

$\begin{array}{l}⇒I=\left(\frac{M}{L}\right)\left[\frac{{x}^{3}}{3}{\right]}_{\frac{-L}{2}}^{\frac{L}{2}}=\frac{M{L}^{2}}{12}\end{array}$

## Mi of a rectangular plate about a line parallel to one of the sides and passing through the center

The figure here shows the small element with repect to the axis of rotation i.e. y-axis, which is parallel to the breadth of the rectangle. Note that axis of rotation is in the place of plate. Here, the steps for calculation are :

(i) Infinetesimally small element of the body :

Let us consider an small element "dx" along the length, which is situated at a linear distance "x" from the axis.

(ii) Elemental mass :

Linear density, λ, is the appropriate density type in this case.

$\begin{array}{l}\lambda =\frac{M}{a}\end{array}$

where "M" and "a" are the mass and length of the rectangular plate respectively. Elemental mass (dm) is, thus, given as :

$\begin{array}{l}đm=\lambda đx=\left(\frac{M}{a}\right)đx\end{array}$

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