18.9 Theorems on moment of inertia (application)  (Page 6/2)

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.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the application of theorem on moment of inertia. For this reason, questions are categorized in terms of the characterizing features pertaining to the questions :

• Theorems on moment of inertia
• Nature of MI
• Part of a rigid body
• Geometric shapes formed from wire

Theorems on moment of inertia

Example 1

Problem : A uniform circular disk of radius "R" lies in "xy" - plane such that its center is at the origin of coordinate system as shown in the figure. Its MI about the axis defined as x = R and y = 0 is equal to its MI about axis defined as y = d and z = 0. Then, find “d”.

Mi of uniform circular disk

Solution : First, we need to visualize the two axes as defined. Then, we find disk's MI about them. Finally, we equate the MIs as given in the question to find "d".

The coordinates x = R and y = 0 define the axis in "xz" - plane as "y" coordinate is zero. Further, the axis intersects x-axis at x = R = a constant. This axis (z'), therefore, is parallel to z-axis as shown in the figure.

Mi of uniform circular disk

Note that z axis passes through center of mass. Thus by applying theorem of parallel axes, the MI of the disk about z' by the theorem of parallel axes is :

$$\begin{array}{l}{I}_{\mathrm{z’}}=I_z+M{R}^2\end{array}$$

$$\begin{array}{l}{I}_{\mathrm{z’}}=\frac{M{R}^2}{2}+M{R}^2=\frac{3M{R}^2}{2}\end{array}$$

The coordinates y = d and z = 0 define the axis in "xy" - plane as "z" coordinate is zero. Further, this axis intersects y-axis at y = d = a constant. This axis (x'), therefore, is parallel to x-axis as shown in the figure.

Mi of uniform circular disk

The MI of the disk about x' by the theorem of parallel axes :

$$\begin{array}{l}{I}_{\mathrm{x’}}=I_x+M{R}^2\end{array}$$

$$\begin{array}{l}{I}_{\mathrm{x’}}=\frac{M{R}^2}{4}+M{d}^2\end{array}$$

According to the question,

$$\begin{array}{l}{I}_{\mathrm{x’}}=I_{\mathrm{z’}}\end{array}$$

$$\begin{array}{l}\frac{M{R}^2}{4}+M{d}^2=\frac{3M{R}^2}{2}\end{array}$$

$$\begin{array}{l}M{d}^2=\frac{3M{R}^2}{2}-\frac{M{R}^2}{4}=\frac{5M{R}^2}{4}\end{array}$$

$$\begin{array}{l}d=\frac{\surd 5R}{2}\end{array}$$

Example 2

Problem : Two identical uniform rods, each of mass "M" and length "L", are joined together to form a cross. Find the MI of the cross about the angle bisector "AA'" as shown in the figure.

Mi of a cross

Solution : MI of each of the rod about perpendicular axis, which passes through COM (O) is :

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

The MI of two rods joined together as a cross about perpendicular axis through the common point is :

$$\begin{array}{l}\Rightarrow I_{\mathrm{cross}}=2I_{\mathrm{rod}}=2x\frac{M{L}^2}{12}=\frac{M{L}^2}{6}\end{array}$$

The cross can be treated as a planar object. As such, we can apply theorem of perpendicular axes to find MI about the bisector of the right angle. We consider another axis BB' perpendicular to AA' in the plane of cross. Then, according to theorem of perpendicular axes,