Moment Of Inertia Of A Rectangular Plate

Moment of Inertia of A Rectangular Plate

$\text { Let } I_{y y}=$ Moment of inertia for uniform rectangular lamina about y-axis passing through its centre.

$\text { To calculate } I_{y y}$

Consider a uniform rectangular lamina of length l, and breadth b and mass M

Mass per unit area of rectangular lamina $=\sigma=\frac{M}{A}=\frac{M}{l \times b}$

Take a small element of mass dm with length dx at a distance x from the y-axis as shown in the figure.

$
\begin{aligned}
& d m=\sigma d A=\sigma(b d x) \\
& \Rightarrow d I=x^2 d m
\end{aligned}
$

Now integrate this dl between the limits $\frac{-l}{2}$ to $\frac{l}{2}$

$I_{y y}=\int d I=\int x^2 d m=\int_{\frac{-1}{2}}^{\frac{1}{2}} \frac{M}{l b} x^2 *(b) d x=\frac{M}{l} \int_{\frac{-1}{2}}^{\frac{1}{2}} x^2 d x=\frac{M l^2}{12}$

Similarly,

Let $I_{xx}$ = Moment of inertia for uniform rectangular lamina about the x-axis passing through its centre.

To calculate $I_{xx}$

Take a small element of mass dm with length dx at a distance x from the x-axis as shown in the figure.

mass per unit Area of rectangular lamina = $\sigma =\frac{M}{A}=\frac{M}{l*b}$

$dm=\sigma dA=\sigma (ldx)$

$\Rightarrow dI= x^2dm$

Now integrate this dI between the limits ${\frac{-b}{2}} \ to \ {\frac{ b}{2}}$

$I_{x x}=\int d I=\int x^2 d m=\int_{\frac{-b}{2}}^{\frac{b}{2}} \frac{M}{l b} x^2 *(l) d x=\frac{M}{b} \int_{\frac{-b}{2}}^{\frac{b}{2}} x^2 d x=\frac{M b^2}{12}$

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Solved Examples Based on A Rectangular Plate

Example 1: If $I=$ Moment of inertia for uniform rectangular lamina of length $l$, and breadth $b$ and mass $M$ about an axis passing through its centre and perpendicular to its breadth. Then I will be equal to:

1) $\frac{M b^2}{12}$
2) $\frac{M l^2}{12}$
3) $\frac{M l^2}{6}$
4) $\frac{M b^2}{6}$

Solution:

From the figure, we can say that

$I_{x x}=I=\frac{M b^2}{12}$

Hence, the correct option is (1).

Example 2: The moment of inertia for a rectangular lamina about the axis in the plane of lamina passing through the end and parallel to the breadth is:

1) $\frac{m l^2}{12}$
2) $\frac{m l^2}{4}$
3) $\frac{m l^2}{3}$
4) $\frac{m l^2}{2}$

Solution

The axis is in the plane of the lamina passing through the end & parallel to the breadth.

The mass distribution of lamina in the above case is like a rod for which the axis passes through its end hence the moment of inertia for a rectangular lamina about the axis in the plane of lamina passing through the end and parallel to the $\text { breadth is } \frac{M l^2}{3} \text { }$

Hence, the answer is the option (3).

Example 3: From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. The moment of inertia of a cube about an axis passing through its centre and perpendicular to one of its faces is :

1) $\frac{M R^2}{32 \sqrt{2} \pi}$
2) $\frac{M R^2}{16 \sqrt{2} \pi}$
3) $\frac{4 M R^2}{9 \sqrt{3} \pi}$
4) $\frac{4 M R^2}{3 \sqrt{3} \pi}$

Solution:

$\begin{aligned}
& a=\frac{2}{\sqrt{ } 3} R \\
& \frac{M}{M^{\prime}}=\frac{\frac{4}{3} \pi R^3}{a^3}=\frac{\frac{4}{3} \pi R^3}{\left(\frac{2}{\sqrt{3}^3} R\right)^3} \Rightarrow \frac{M}{M^{\prime}}=\frac{\frac{4}{3} \pi R^3}{\frac{8}{3 \sqrt{3}} R^3}=\frac{4 \pi}{3} \times \frac{3 \sqrt{3}}{8} \\
& \frac{M}{M^{\prime}}=\frac{\sqrt{3} \pi}{2} \Rightarrow M^{\prime}=\frac{2 M}{\sqrt{3} \pi}
\end{aligned}$

$\therefore$ M.O.I. of the cube about the given axis.

$I=\frac{M^{\prime} a^2}{6}=\frac{\frac{2 M}{\sqrt{3} \pi} \times\left(\frac{2}{\sqrt{3}} R\right)^2}{6}=\frac{4 M R^2}{9 \sqrt{3} \pi}$

Hence, the answer is the option (3).

Example 4: For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point ) is:

1) $\frac{2}{3}$
2) $\frac{1}{4}$
3) $\frac{1}{8}$
4) $\frac{1}{2}$

Solution:

$\begin{aligned}
& \mathrm{I}_{\mathrm{O}}=\frac{\mathrm{M}}{12}\left[\mathrm{~L}^2+\mathrm{B}^2\right]=\frac{\mathrm{M}}{12}\left[80^2+60^2\right] \\
& \mathrm{I}_{\mathrm{O}^{\prime}}=\mathrm{I}_0+\mathrm{Md}^2\{\text { parallel axis theorem } \\
& =\frac{\mathrm{M}}{12}\left[80^2+60^2\right]+\mathrm{M}[50]^2 \\
& \frac{\mathrm{I}_{\mathrm{O}}}{\mathrm{I}_{\mathrm{O}^{\prime}}}=\frac{\mathrm{M} / 12\left[80^2+60^2\right]}{\frac{\mathrm{M}}{2}\left[80^2+60^2\right]+\mathrm{M}\left[50^2\right.}=\frac{1}{4}
\end{aligned}$

Hence, the answer is the option (4).

Example 5: The moment of inertia of a square plate of side $l$ about the axis passing through one of the corners and perpendicular to the plane of the square plate is given by:

1) $\frac{\mathrm{M} l^2}{6}$
2) $\frac{2}{3} \mathrm{M} l^2$
3) $\mathrm{Ml}^2$
4) $\frac{M l^2}{12}$

Solution:

$\begin{aligned}
& I_{c o M}=\frac{M}{12}\left(L^2+L^2\right)=\frac{M L^2}{6} \\
& I_0=I_{C O M}+M d^2 \\
& \quad=\frac{M L^2}{6}+M\left(\frac{L}{\sqrt{2}}\right)^2=\frac{M L^2}{6}+\frac{M L^2}{2} \\
& I_0=\frac{4 M L^2}{6}=\frac{2}{3} M L^2
\end{aligned}$

Summary

The moment of inertia is applied in both linear and angular moments, although it manifests itself in planar and spatial movement in rather different ways. We got you the derivations of how to calculate the moment of inertia of a rectangular plate, along with the derivation of the moment of inertia of a rectangular plate about its centre, and the moment of inertia of a rectangular plate about its edge as well.