18.7 Moments of inertia of rigid bodies (application)  (Page 2/3)

$$\begin{array}{l}{I}_s=\frac{2M{R_s}^2}{5}\end{array}$$

Let the radius of the ring is $R_r$ . The MI of the ring about its central axis is given as :

$$\begin{array}{l}{I}_r=M{R_r}^2\end{array}$$

According to question, the moments of inertia of the solid sphere and the ring about their central axes are same :

$$\begin{array}{l}M{R_r}^2=\frac{2M{R_s}^2}{5}\end{array}$$

$$\begin{array}{l}\Rightarrow R_r=\surd \left(\frac{2}{5}\right)x{R}_s\end{array}$$

Estimation of mi by inspection

Example 3

Problem : Find the axis of the coordinate system about which the moment of inertia of a uniform rectangle of dimensions "a" and "2a" is least.

Rectangular plate

Solution : We can actually determine MI about the given axes by evaluating the integral of MI about each of the axes and then compare. However, we can as well estimate least MI as mass of the rectangle is uniformly distributed. MI is least for the axis about which the mass is distributed closest to it.

Rectangular plate

We can see here that the most distant mass about x-axis is at a distance "a/2"; most distant mass about y-axis is at a distance "a"; most distant mass about z-axis at a distance between "a" and "√5a/2".

Here, mass distribution is closest about x-axis. Thus, moment of inertia about x-axis is least.

Part of a rigid body

Example 4

Problem : Find the the MI of one quarter of a circular plate of mass "M" and radius "R" about z-axis, as shown in the figure.

Quarter plate

Solution : We need to look closely the way MI about an axis is defined. MI of the rigid body about the an axis is given by :

$$\begin{array}{l}I=\sum M_i{R_i}^2\end{array}$$

This definition is essentially scalar in nature. The "R" in the expression is perpendicular distance of a particle from the axis of rotation. Does it matter whether the particle lie on left or right of the axis? Obviously no. It means that MI has no directional attribute. Thus, we can conclude that MI of the quarter plate is arithmetic quarter (1/4) of the MI of a complete disk, whose mass is 4 times greater than that of quarter plate.

Circular plate

Hence, mass of the corresponding complete disk is 4M and radius is same R as that of quarter plate. The MI of the complete disk about perpendicular axis to its surface is :

$$\begin{array}{l}{I}_O=\left(4M\right)\frac{R^2}{2}\end{array}$$

Now, MI of the quarter plate is 1/4 of the complete disk,

$$\begin{array}{l}I=\frac{1}{4}{I}_O=\frac{1}{4}\left(4M\right)\frac{R^2}{2}=\frac{M{R}^2}{2}\end{array}$$

Note : We notice that the expression of MI has not changed for the quarter plate. We must, however, be aware that the symbol "M" represents mass of the quarter plate - not that of the complete circular plate.

QBA (Question based on above) : The MI of a uniform disk is 1.0 $\mathrm{kg}-m^2$ about its perpendicular central axis. If a segment, subtending an angle of 120° at the center, is removed from it, then find the MI of the remaining disk about the axis.

Hint : We see that 120°/360° = 1/3 part of the complete disk is removed. Therefore, the mass of the remaining part is 1-1/3 = 2/3. Answer is “2/3 $\mathrm{kg}-m^2$ ”.

Oblique axis

Example 5

Problem : Determine the moment of inertia of a uniform rod of length "L" and mass "m" about an axis passing through its center and inclined at an angle "θ" as shown in the figure.

Mi of a rod