Connexions module: m14292 1
Moments of inertia of rigid bodies∗
Sunil Kumar Singh
This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License †
Abstract Moment of inertia of rigid body depends on the distribution of mass about the axis of rotation. In the module titled Rotation of rigid body 1, we derived expressions of moments of inertia (MI) for dierent object forms as :
1. For a particle : I = mr2
P 2 2. For a system of particles : I = miri
3. For a rigid body : I = R r2m In this module, we shall evalaute MI of dierent regularly shaped rigid bodies.
1 Evaluation strategy We evaluate right hand integral of the expression of moment of inertia for regularly shaped geometric bodies. The evaluation is basically an integration process, well suited to an axis of rotation for which mass distribution is symmetric. In other words, evaluation of the integral is easy in cases where mass of the body is evenly distributed about the axis. This axis of symmetry passes through "center of mass" of the regular body. Calculation of moment of inertia with respect to other axes is also possible, but then integration process becomes tedious. There are two very useful theorems that enable us to calculate moment of inertia about certain other relevant axes as well. These theorems pertaining to calculation of moment of inertia with respect to other relevant axes are basically "short cuts" to avoid lengthy integration. We must, however, be aware that these theorems are valid for certain relevant axes only. If we are required to calculate moment of inertia about an axis which can not be addressed by these theorems, then we are left with no choice other than evaluating the integral or determining the same experimentally. As such, we limit ourselves in using integral method to cases, where moment of inertia is required to be calculated about the axis of symmetry. In this module, we will discuss calculation of moment of inertia using basic integral method only, involving bodies having (i) regular geometric shape (ii) uniform mass distribution i.e uniform density and (iii) axis of rotation passing through center of mass (COM). Application of the theorems shall be discussed in a separate module titled " Theorems on moment of inertia 2".
∗Version 1.10: May 23, 2011 2:48 am GMT-5 †http://creativecommons.org/licenses/by/3.0/ 1"Rotation of rigid body"
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As far as integration method is concerned, it is always useful to have a well planned methodology to complete the evaluation. In general, we complete the integration in following steps : 1. Identify an innitesimally small element of the body. 2. Identify applicable density type (linear, surface or volumetric). Calculate elemental mass "dm" in terms of appropriate density. 3. Write down the expression of moment of inertia (I) for elemental mass. 4. Evaluate the integral of moment of inertia for an appropriate pair of limits and determine moment of inertia of the rigid body. Identication of small element is crucial in the evaluation of the integral. We consider linear element in evaluating integral for a linear mass distribution as for a rod or a plate. On the other hand, we consider thin concentric ring as the element for a circular plate, because we can think circular plate being composed of innite numbers of thin concentric rings. Similarly, we consider a spherical body, being composed of closely packed thin spherical shells. Calculation of elemental mass "m" makes use of appropriate density on the basis of the nature of mass distribution in the rigid body :
mass = appropriate density x geometric dimension 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 dierent cases are :
m = λ x m = σ A (1) m = ρ V
Elemental mass
(a) (b)
Figure 1: (a) Elemental mass of a rod for MI about perpendicular bisector (b) Elemental mass of a circular disc for MI about perpendicular axis through the center
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Elemental mass
Figure 2: Elemental mass of a sphere for MI about a diameter
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.
2 Evaluation of moment of inertia In this section, we shall determine MI of known geometric bodies about the axis of its symmetry.
2.1 MI of a uniform rod about its perpendicular bisector The gure here shows the small element with repect to the axis of rotation. Here, the steps for calculation are :
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Moment of inertia
Figure 3: MI of a rod about perpendicular bisector
(i) Innetesimally 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.
M λ = L where "M" and "L" are the mass and length of the rod respectively. Elemental mass (m) is, thus, given as :