Question

1) A uniform square plate S (side c) and a uniform rectangular plate R (sides b, c) have identical areas and masses (figure)

Show that:

(a) $\frac{I_{xR}}{I_{xS}}<1$

2) A uniform square plate S (side c) and a uniform rectangular plate R (sides b, c) have identical areas and masses $\left(figure\right)$

Show that:

(b) $\frac{I_{yR}}{I_{yS}}>1$


3) A uniform square plate S (side c) and a uniform rectangular plate R (sides b, c) have identical areas and masses (figure)

Show that:

(c) ${\frac{I_{zR}}{I}}_{zS}>1$

Hint: Perpendicular axis theorem.

Answer 1

(A) Formul used : $I=\frac{m{r}^2}{12}$

Step 1: Draw a diagram of plate.


Step : 2 Calculate moment of inertia of a plate about x-axis.

MOI of rectangular plate about x-axis, $I=\frac{{}^m{R}^{b^2}}{12}$

MOI of square plate about x-axis, $I=\frac{{}^m{S}^{c^2}}{12}$

Given, $m_R=m_S=m$

Area of surface = Area of rectangular plate

$C^2=ab$

$\frac{I_{xR}}{I_{x}S}=\frac{\frac{m{b}^2}{12}}{\frac{m{c}^2}{12}}=\frac{b^2}{c^2}$

From diagram, $c>b,c^2>b^2$

$\therefore \frac{b^2}{c^2}<1or{\left(\frac{b}{c}\right)}^2<1or{I}_{xR}<I_{xS}$

Final Answer: $\frac{I_{x}R}{I_{xS}}<1$

(B) Formula used: $I=\frac{m{r}^2}{12}$

Step 1: Draw a rough diagram of plate.


Step 2: Calculate moment of inertia of a plate about y-axis.

MOI of rectangular plate about y-axis, $I_{yR}=\frac{m{a}^2}{12}$

MOI of square plate about y-axis, $I_{yS}=\frac{m{c}^2}{12}$


$\frac{I_{yR}}{I_{yS}}=\frac{M{a}^2}{12}\times \frac{12}{M{c}^2}=\frac{a^2}{a^2}$


$\frac{I_{xR}}{I_{xS}}\times \frac{I_{yR}}{I_{yS}}=\frac{b^2}{c^2}\times \frac{a^2}{c^2}=\frac{a^2{b}^2}{c^4}=1$

Since $\frac{I_{xR}}{I_{xS}}<1\Rightarrow \frac{I_{yR}}{I_{yS}}>1$

Hence, $\frac{I_{yR}}{I_{yS}}>1$

Final Answer: $\frac{I_{yR}}{I_{yS}}>1$

C) Formula used: Moment of inertia of a plate is given by $I=\frac{m{l}^2}{12}$

From the perpendicular axis theorem,

$I_z=I_x+I_y$

Moment of inertia of square plate about z-axis, $I_{zS}=\frac{m{c}^2}{12}+=\frac{m{c}^2}{12}=\frac{m{c}^2}{6}$

Moment of inertia of rectangular plate about z-axis, $I_{zR}=\frac{m{b}^2}{12}+\frac{m{a}^2}{12}$

$\frac{I_{z}R}{IzS}=\frac{m(c^2+b^2)}{12}\times \frac{6}{m{c}^2}=(a^2+b^2)/2c^2$

$\because ab=c^2$

$\frac{(a^2+b^2)}{2ab}=\frac{(a^2+b^2-2ab+2ab)}{2ab}$

$\frac{IzR}{IzS}=\frac{(a-b{)}^2+2ab}{2ab}$

$\frac{IzR}{IzS}=\frac{(a-b{)}^2}{2ab}+>1$

$\frac{IzR}{IzS}>1$

Final Answer: $\frac{IzR}{IzS}>1$