18.7 Moments of inertia of rigid bodies (application)  (Page 12/3)

Questions and their answers are presented here in the module text format as if it were an extension of the theoretical treatment of the topic. The idea is to provide a verbose explanation of the solution, detailing the application of theory. Solution presented here, therefore, is treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Hints on solving problems

• Moment of inertia is a scalar quantity without any directional property. This has important implication in solving problems, which involve removal of a part of the rigid body from the whole. We need to simply use the MI formula for a given shape with changed mass of the remaining body, ensuring that the pattern of mass distribution about the axis has not changed.
• Moment of inertia about an oblique axis involves MI integration with certain modification. Here, mass distribution and distance involve different variables. We are required to express integral in terms of one variable so that the same can be integrated by single integration process.
• In determining relative MIs, we should look how closely or how distantly mass is distributed. This enables us to compare MIs without actual calculation in some cases.
• So far, we have studied calculation of MI for uniform objects. However, we can also evaluate MI integral, if variation in mass follows certain pattern and the same can be expressed in terms of mathematical expression involving variable.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the calculation of moment of inertia of regularly shaped rigid bodies. For this reason, questions are categorized in terms of the characterizing features pertaining to the questions as :

• Using MI formula
• Estimation of MI by inspection
• Part of a rigid body
• Non-uniform mass distribution

Using mi formula

Example 1

Problem : The moment of inertia of a straight wire about its perpendicular bisector and moment of inertia of a circular frame about its perpendicular central axis are $I_1$ and $I_2$ respectively. If the composition of wires are same and lengths of the wires in them are equal, then find the ratio $\raisebox{1ex}{$I_1$}\!\left/ \!\raisebox{-1ex}{$I_2$}\right.$ .

Solution : The MI of the straight wire about a perpendicular through the mid point is :

$$\begin{array}{l}{I}_1=\frac{M{L}^2}{12}\end{array}$$

The MI of the circular frame about its perpendicular central axis is :

$$\begin{array}{l}{I}_2=M{R}^2\end{array}$$

According to question, the lengths of the wires in them are equal. As the composition of the wires same, the mass of two entities are same. Also :

$$\begin{array}{l}L=2\pi R\\ \Rightarrow R=\frac{L}{2\pi }\end{array}$$

Substituting in the expression of MI of circular frame,

$$\begin{array}{l}{I}_2=\frac{M{L}^2}{4{\pi }^2}\end{array}$$

Now, the required ratio is :

$$\begin{array}{l}\frac{I_1}{I_2}=\frac{M{L}^{2}x4{\pi }^2}{12xM{L}^2}\end{array}$$

$$\begin{array}{l}\frac{I_1}{I_2}=\frac{{\pi }^2}{3}\end{array}$$

Example 2

Problem : The moments of inertia of a solid sphere and a ring of same mass about their central axes are same. If $R_s$ be the radius of solid sphere, then find the radius of the ring.

Solution : The MI of the solid sphere about its central axis is given as :