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Let us consider the small mass of volume "dV" of the sphere of thickness "dr", which is situated at a linear distance "r" from the center of the sphere.
(ii) Elemental mass :
Volumetric 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 uniform solid sphere respectively. Here, hollow sphere of infinitesimal thickness itself is considered as elemental mass (dm) :
$$\begin{array}{l}\u0111m=\rho \u0111V=\left(\frac{M}{\frac{4}{3}\pi {R}^{3}}\right)4\pi {r}^{2}\u0111r\end{array}$$
(iii) Moment of inertia for elemental mass
Moment of inertia of elemental mass is obtained by using expresion of MI of hollow sphere :
$$\begin{array}{l}\u0111I=\frac{2}{3}{r}^{2}\u0111m\end{array}$$
$$\begin{array}{l}\Rightarrow \u0111I=\frac{2}{3}{r}^{2}\left(\frac{3M}{{R}^{3}}\right){r}^{2}\u0111r=\left(\frac{2M}{{R}^{3}}\right){r}^{4}\u0111r\end{array}$$
(iv) Moment of inertia of rigid body
$$\begin{array}{l}I=\int \left(\frac{2M}{{R}^{3}}\right){r}^{4}\u0111r\end{array}$$
Taking out the constants from the integral sign, we have :
$$\begin{array}{l}I=\left(\frac{2M}{{R}^{3}}\right)\int {r}^{4}\u0111r\end{array}$$
The appropriate limits of integral in this case are 0 and R. Hence,
$$\begin{array}{l}\Rightarrow I=\left(\frac{2M}{{R}^{3}}\right){\int}_{0}^{R}{r}^{4}\u0111r\end{array}$$
An inspection of the expressions of MIs of different bodies reveals that it is directly proportional to mass of the body. It means that a heavier body will require greater torque to initiate rotation or to change angular velocity of rotating body. For this reason, the wheel of the railway car is made heavy so that it is easier to maintain speed on the track.
Further, MI is directly proportional to the square of the radius of circular object (objects having radius). This indicates that geometric dimensions of the body have profound effect on MI and thereby on rotational inertia of the body to external torque. Engineers can take advantage of this fact as they can design rotating part of a given mass to have different MIs by appropriately distributing mass either closer to the axis or away from it.
The MIs of ring, disk, hollow cylinder, solid cylinder, hollow sphere and solid sphere about their central axes are $M{R}^{2},\phantom{\rule{2pt}{0ex}}\frac{M{R}^{2}}{2},\phantom{\rule{2pt}{0ex}}M{R}^{2},\phantom{\rule{2pt}{0ex}}\frac{2M{R}^{2}}{3}$ and $\frac{2M{R}^{2}}{5}$ respectively. Among these, the MIs of a ring and hollow cylinder are $M{R}^{2}$ as all mass elements are distributed at equal distance from the central axis. For other geometric bodies that we have considered, we find thaheir MIs expressions involve some fraction of $M{R}^{2}$ as mass distribution is not equidistant from the central axis.
We can, however, assume each of other geometric bodies (not necessarily involving radius) equivalent to a ring (i.e a hoop) of same mass and a radius known as "radius of gyration". In that case, MI of a rigid body can be written in terms of an equivalent ring as :
Formally, we can define radius of gyration as the radius of an equivalent ring of same moment of inertia. In the case of solid sphere, the MI about one of the diameters is :
$$\begin{array}{l}I=\frac{2M{R}^{2}}{5}\end{array}$$
By comparison, the radius of gyration of solid sphere about its diameter is :
$$\begin{array}{l}K=\surd \left(\frac{2}{5}\right)xR\end{array}$$
Evidently, radius of gyration of a rigid body is specific to a given axis of rotation. For example, MI of solid sphere about an axis parallel to its diameter and touching its surface (we shall determine MI about parallel axis in the next module) is :
$$\begin{array}{l}I=\frac{7M{R}^{2}}{5}\end{array}$$
Thus, radius of gyration of solid sphere about this parallel axis is :
$$\begin{array}{l}K=\surd \left(\frac{7}{5}\right)xR\end{array}$$
The radius of gyration is a general concept for rotational motion of a rigid body like moment of inertia and is not limited to circularly shaped bodies, involving radius. For example, radius of gyration of a rectangular plate about a perpendicular axis passing through its COM and parallel to one of its breadth (length “a” and breadth “b”) is :
$$\begin{array}{l}I=\frac{M{a}^{2}}{12}\end{array}$$
and the radius of gyration about the axis is :
$$\begin{array}{l}K=\frac{a}{\surd 12}\end{array}$$
1. The results of some of the known geometric bodies about their central axes, as derived in this module, are given in the figures below :
2. The radius of gyration (K) is defined as the radius of an equivalent ring of same moment of inertia. The radius of gyration is defined for a given axis of rotation and has the unit that of length. The moment of inertia of a rigid body about axis of rotation, in terms of radius of gyration, is expressed as :
$$\begin{array}{l}I=M{K}^{2}\end{array}$$
3. Some other MIs of interesting bodies are given here. These results can be derived as well in the same fashion with suitably applying the limits of integration.
(i)The MI of a thick walled cylinder ( ${R}_{1}$ and ${R}_{2}$ are the inner and outer radii)
$$\begin{array}{l}I=\frac{M({R}_{1}^{2}+{R}_{2}^{2})}{2}\end{array}$$
(ii) The MI of a thick walled hollow sphere ( ${R}_{1}$ and ${R}_{2}$ are the inner and outer radii )
$$\begin{array}{l}I=\frac{2}{5}xMx\frac{({R}_{2}^{5}-{R}_{1}^{5})}{({R}_{2}^{3}-{R}_{1}^{3})}\end{array}$$
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