18.6 Moments of inertia of rigid bodies  (Page 3/5)

(iii) Moment of inertia for elemental mass :

Moment of inertia of elemental mass is :

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

(iv) Moment of inertia of rigid body :

Proceeding in the same manner as for the case of an uniform rod, the MI of the plate about the axis is given by :

Similarly, we can also calculate MI of the rectangular plate about a line parallel to its length and through the center,

Mi of a circular ring about a perpendicular line passing through the center

The figure here shows the small element with respect to the axis of rotation. Here, we can avoid the steps for calculation as all elemental masses are at a fixed (constant) distance "R" from the axis. This enables us to take "R" directly out of the basic integral :

Moment of inertia

We must note here that it was not possible so in the case of a rod or a rectangular plate as elemental mass is at a variable distance from the axis of rotation. We must realize that simplification of evaluation in this case results from the fact that masses are at the same distance from the axis of rotation. This means that the expression of MI will be valid only when thickness of the ring is relatively very small in comparison with its radius.

Mi of a thin circular plate about a perpendicular line passing through the center

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

Moment of inertia

(i) Infinetesimally small element of the body :

Let us consider an small circular ring element of width "dr", which is situated at a linear distance "r" from the axis.

(ii) Elemental mass :

Areal 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 area of the plate respectively. Here, ring of infinitesimal thickness itself is considered as elemental mass (dm) :

$$\begin{array}{l}đm=\sigma đA=\left(\frac{M}{\pi R^2}\right)2\pi rđr\\ đm=\frac{2Mrđr}{R^2}\end{array}$$

(iii) Moment of inertia for elemental mass

MI of the ring, 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 the constants out of the integral sign, we have :

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

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

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

Moment of inertia of a hollow cylinder about its axis

The figure here shows the small element with repect to the axis of rotation. Here, we can avoid the steps for calculation as all elemental masses composing the cylinder are at a fixed (constant) distance "R" from the axis. This enables us to take "R" out of the integral :

Moment of inertia

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

Moment of inertia of a uniform solid cylinder about its longitudinal axis

The figure here shows the small element as hollow cylinder with repect to the axis of rotation. We must note here that we consider hollow cylinder itself as small element for which the MI expression is known. Here, the steps for calculation are :