18.6 Moments of inertia of rigid bodies  (Page 4/5)

Moment of inertia

(i) Infinetesimally small element of the body :

Let us consider the small mass of volume "dV" of the hollow cylinder of small thickness "dr" in radial direction, which is situated at a linear distance "r" from the axis.

(ii) Elemental mass :

Volume density, ρ, is the appropriate density type in this case.

$$\begin{array}{l}\rho =\frac{M}{V}\end{array}$$

where "M" and "V" are the mass and volume of the solid cylinder respectively. Here, cylinder of infinitesimal thickness itself is considered as elemental mass (dm) :

$$\begin{array}{l}đm=\rho đV=\left(\frac{M}{V}\right)đV=\left(\frac{M}{\pi R^{2}L}\right)2\pi rLđr\\ đm=\frac{2Mrđr}{R^2}\end{array}$$

(iii) Moment of inertia for elemental mass

MI of the hollow cylinder, which is treated as elemental mass, is given by :

$$\begin{array}{l}đI=r^2đm=r^2\frac{2Mrđr}{R^2}\end{array}$$

(iv) Moment of inertia of rigid body

$$\begin{array}{l}I=\int r^2đm=\int \frac{2r^{2}Mrđr}{R^2}\end{array}$$

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

$$\begin{array}{l}I=\left(\frac{2M}{R^2}\right)\int r^3đr\end{array}$$

The appropriate limita of integral in this case are 0 and R. Hence,

$$\begin{array}{l}I=\left(\frac{2M}{R^2}\right){\int }_0^R{r}^3đr\end{array}$$

We note here that MI of solid cyliner about its longitudinal axis is same as that of a plate. Another important aspect of MI, here, is that it is independent of the length of solid cylinder.

Moment of inertia of a hollow sphere about a diameter

The figure here shows that hollow sphere can be considered to be composed of infinite numbers of rings of variable radius. Let us consider one such ring as the small element, which is situated at a linear distance "R" from the center of the sphere. The ring is positioned at an angle "θ" with respect to axis of rotation i.e. y-axis as shown in the figure.

Moment of inertia

(i) Infinetesimally small element of the body :

Let us consider the small mass of area "dA" of the ring of small thickness "Rdθ", which is situated at a linear distance "R" from the center of the sphere.

(ii) Elemental mass :

Area density, σ, is the appropriate density type in this case.

$$\begin{array}{l}\sigma =\frac{M}{A}\end{array}$$

where "M" and "A" are the mass and surface area of the hollow sphere respectively. Here, ring of infinitesimal thickness itself is considered as elemental mass (dm) :

$$\begin{array}{l}đm=\sigma đA=\left(\frac{M}{4\pi R^2}\right)2\pi r\sin \theta Rđ\theta \\ đm=\left(\frac{M}{2}\right)\sin \theta đ\theta \end{array}$$

(iii) Moment of inertia for elemental mass

Here, radius of elemental ring about the axis is R sinθ. Moment of inertia of elemental mass is :

$$\begin{array}{l}đI=R^2{\sin }^2\theta đm=R^2{\sin }^2\theta \left(\frac{M}{2}\right)\sin \theta đ\theta \end{array}$$

(iv) Moment of inertia of rigid body

$$\begin{array}{l}I=\int \left(\frac{M}{2}\right)R^2{\sin }^2\theta đ\theta \end{array}$$

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

$$\begin{array}{l}I=\left(\frac{M{R}^2}{2}\right)\int {\sin }^3\theta đ\theta \end{array}$$

The appropriate limits of integral in this case are 0 and π. Hence,

$$\begin{array}{l}\Rightarrow I=\left(\frac{M{R}^2}{2}\right){\int }_0^{\pi }{\sin }^3\theta đ\theta \end{array}$$

$$\begin{array}{l}\Rightarrow I=\left(\frac{M{R}^2}{2}\right){\int }_0^{\pi }(1-{\cos }^2\theta )\sin \theta đ\theta \end{array}$$

$$\begin{array}{l}\Rightarrow I=\left(\frac{M{R}^2}{2}\right){\int }_0^{\pi }-(1-{\cos }^2\theta )đ\left(\cos \theta \right)\end{array}$$

Moment of inertia of a uniform solid sphere about a diameter

A solid sphere can be considered to be composed of concentric spherical shell (hollow spheres) of infinitesimally small thickness "dr". We consider one hollow sphere of thickness "dr" as the small element, which is situated at a linear distance "r" from the center of the sphere.

Moment of inertia

(i) Infinetesimally small element of the body :