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

Mi of a cross about angle bisector

$$\begin{array}{l}{I}_{\mathrm{cross}}=I_{\mathrm{AA’}}+I_{\mathrm{BB’}}\end{array}$$

The MIs about the two bisectors are equal by the symmetry of the cross about them. Hence,

$$\begin{array}{l}{I}_{\mathrm{AA’}}=I_{\mathrm{BB’}}\end{array}$$

Thus,

$$\begin{array}{l}\Rightarrow I_{\mathrm{cross}}=I_{\mathrm{AA’}}+I_{\mathrm{BB’}}=2I_{\mathrm{AA’}}\end{array}$$

$$\begin{array}{l}\Rightarrow I_{\mathrm{AA’}}=\frac{I_{\mathrm{cross}}}{2}=\frac{M{L}^2}{12}\end{array}$$

Nature of mi

Example 3

Problem : The MI of a solid sphere is calculated about an axis parallel to one of its diameter, at a distance "x" from the origin. The values are plotted against distance. Which of the four plots, as shown below, correctly describes the variation of MI with respect to "x" for the solid sphere.

Plots of mi .vs. distance

Solution : It is evident that farther the axis from the origin, greater is MI about that axis. We can, thus, conclude that the plot between MI and the distance should be increasing one with increasing distance ("x"). Thus, options (a) and (d) are incorrect. Now, the MI of solid sphere about an axis parallel to one of diameters, at a distance "x" from the origin is given as :

Plots of mi .vs. distance

$$\begin{array}{l}I=I_O+M{x}^2\end{array}$$

where $I_O$ is MI about one of the diameters and is a constant for the given solid sphere. Thus, the equation above takes the form of the equation of parabola,

$$\begin{array}{l}y=m{x}^2+c\end{array}$$

The correct plot, therefore, is an increasing parabola as shown in Figure (c).

Part of a rigid body

Example 4

Problem : Four holes of radius "L/4" are made in a thin square plate of mass "M" and side "L" in "xy" - plane, as shown in the figure. Find its MI about z – axis.

Mi of remaining body

Solution : MI is a scalar quantity. It means that MI has no directional attribute. Thus, we can conclude that MI of the complete plate is equal to the sum of MIs of remaining plate and that of the four circular disks taken out from the plate about z-axis. Hence, MI of remaining plate is obtained as :

$$\begin{array}{l}\text{I (MI of remaining plate)}=\text{Iz (MI of complete plate)}-4x\text{Icz (MI of circular disks)}\end{array}$$

Now, according to theorem of perpendicular axes, the MI of complete square about z-axis is two times its MI about x-axis :

$$\begin{array}{l}\Rightarrow I_z=2x\frac{M{L}^2}{12}=\frac{M{L}^2}{6}\end{array}$$

Further, according to theorem of parallel axes, the MI of circular disk of mass “m” and radius “L/4” about the same z-axis is :

Mi of remaining body

$$\begin{array}{l}\Rightarrow I_{\mathrm{cz}}=m{\left(\frac{L}{4}\right)}^2+m{\left(\frac{\surd 2L}{4}\right)}^2\end{array}$$

$$\begin{array}{l}\Rightarrow I_{\mathrm{cz}}=\frac{m{L}^2}{16}+\frac{2m{L}^2}{16}=\frac{3m{L}^2}{16}\end{array}$$

The mass of the disk "m" is given by multiplying areal density with the area of the circular disk,

$$\begin{array}{l}m=\frac{M}{L^2}x\pi {\left(\frac{L}{4}\right)}^2=\frac{\pi M}{16}\end{array}$$

Substituting for "m", we have :

$$\begin{array}{l}\Rightarrow I_{\mathrm{cz}}=\frac{3m{L}^2}{16}=\frac{3\pi M{L}^2}{16x16}\end{array}$$

Thus, the required MI is :

$$\begin{array}{l}\Rightarrow I=\frac{M{L}^2}{6}-4x\frac{3\pi M{L}^2}{16x16}=\frac{16x16M{L}^2-12x6\pi M{L}^2}{6x16x16}\end{array}$$

$$\begin{array}{l}\Rightarrow I=\frac{32M{L}^2-9\pi M{L}^2}{\mathrm{192}}\end{array}$$

Geometric shapes formed from wire

Example 5

Problem : A thin wire of length "L" and uniform linear mass density "λ" is bent into a circular loop. Find the MI of the loop about an axis AA' as shown in the figure.

Mi of a circular loop

Solution : The MI of the circular loop about a tangential axis AA' is found out by the theorem of parallel axis :

Mi of a circular loop

$$\begin{array}{l}{I}_{\mathrm{AA’}}=I_{\mathrm{XX’}}+M{R}^2\end{array}$$

where IXX' is the MI about an axis which is one of the diameters of the ring and is parallel to tangential axis AA'. Now, MI about the diameter can be calculated applying theorem of perpendicular axes :

$$\begin{array}{l}\Rightarrow I_{\mathrm{XX’}}=\frac{I_O}{2}\end{array}$$

where $I_O$ is MI about an axis perpendicular to the surface of loop and passing through center of mass (ZZ' axis). It is given by the expression :

Mi of a circular loop

$$\begin{array}{l}\Rightarrow I_O=M{R}^2\end{array}$$

Combining two equations, we have :

$$\begin{array}{l}\Rightarrow I_{\mathrm{XX’}}=\frac{M{R}^2}{2}\end{array}$$

Putting this in the expression of $I_{\mathrm{AA\text{'}}}$ , we have :

$$\begin{array}{l}\Rightarrow I_{\mathrm{AA’}}=\frac{M{R}^2}{2}+M{R}^2=\frac{3M{R}^2}{2}\end{array}$$

Now, we are required to find mass and radius from the given data. Here,

$$\begin{array}{l}M=\lambda L\end{array}$$

Since the wire is bent into a circle, the perimeter of the circle is equal to the length of the wire. Hence,

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

Putting these values in the expression of MI about AA' axis is :

$$\begin{array}{l}\Rightarrow I_{\mathrm{AA’}}=\frac{3M{R}^2}{2}=\frac{3\lambda Lx{L}^2}{2{\left(2\pi \right)}^2}=\frac{3\lambda L^3}{8{\pi }^2}\end{array}$$

Example 6

Problem : A uniform wire of mass "M" and length "L" is bent into a regular hexagonal loop. Find its MI about an axis passing through center of mass and perpendicular to its surface.

Solution : We find MI of hexagonal loop in following three steps :

1. Treat each side as a wire of mass M/6 and length L/6 and find its MI about perpendicular axis through center of mass of the side.
2. Apply theorem of parallel axes to find MI of the side about perpendicular axis through center of mass of the loop.
3. Add MIs of individual sides to find the MI of the loop about perpendicular axis through center of mass.

The MI of the side about perpendicular axis through center of mass of the side is :

Mi of a regular hexagonal loop

$$\begin{array}{l}\Rightarrow I_{\mathrm{AA’}}=\frac{\frac{M}{6}x{\left(\frac{L}{6}\right)}^2}{12}=\frac{M{L}^2}{12x{6}^3}\end{array}$$

By the geometry of a side with respect to center, the traingle OPQ is a right angle triangle. We have perpendicular distance "R" between axes AA' and ZZ' as :

Mi of a side of regular hexagonal loop

$$\begin{array}{l}\Rightarrow \tan 60°=\frac{R}{\frac{L}{12}}\end{array}$$

$$\begin{array}{l}\Rightarrow R=\frac{L}{12}x\surd 3=\frac{\surd 3}{12}L\end{array}$$

Applying theorem of parallel axes, the MI about perpendicular axis through center of mass of the loop is :

$$\begin{array}{l}{I}_O=I_{\mathrm{zz’}}=I_{\mathrm{AA’}}+\frac{M}{6}x{R}^2\\ \Rightarrow I_O=\frac{M{L}^2}{12x{6}^3}+\frac{M}{6}x{\left(\frac{\surd 3L}{12}\right)}^2\\ \Rightarrow I_O=\frac{M{L}^2}{12x{6}^3}+\frac{3M{L}^2}{{12}^{2}x6}\end{array}$$

$$\begin{array}{l}\Rightarrow I_O=\frac{M{L}^2+9M{L}^2}{12x{6}^3}=\frac{10M{L}^2}{12x{6}^3}\end{array}$$

Now, adding MIs of each side, the required MI about perpendicular axis through center of mass of the loop is :

$$\begin{array}{l}\Rightarrow I=6I_O=\frac{6x10M{L}^2}{12x{6}^3}=\frac{5M{L}^2}{216}\end{array}$$