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A uniform square plate S and a uniform rectangular plate R have identical areas and masses. Show that

$a) \frac{IxR}{IxS}<1$

$b) \frac{IyR}{IyS}>1$

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

Answers (1)

$m_{R}=m_{s}=m$

Area of square = Area of Rectangle

$c^{2}=ab$

$I=mr^{2}$

$\frac{I_{xR}}{I_{xZ}}=\frac{m\left ( \frac{b}{2} \right )^{2}}{m\left ( \frac{c}{2} \right )^{2}}=\frac{b^{2}}{4}\frac{4}{c^{2}}=\frac{b^{2}}{c^{2}}$

$c>b$

$c^{2}>b^{2}$

$1>\frac{b^{2}}{c^{2}}$

Thus $\frac{I_{xR}}{I_{xZ}}<1$

(b) $\frac{I_{yR}}{I_{yZ}}=\frac{m\left ( \frac{a}{2} \right )^{2}}{m\left ( \frac{c}{2} \right )^{2}}=\frac{a^{2}}{4}=\frac{4}{c^{2}}=\frac{a^{2}}{c^{2}}$

$a>c$

$\frac{a^{2}}{c^{2}}>1$

$\frac{I_{yR}}{I_{yZ}}>1$

(c) $I_{zR}-I_{zS}=m\left ( \frac{d_{R}}{2} \right )^{2}-m\left ( \frac{d_{s}}{2} \right )^{2}$

$I_{zR}-I_{zS}=\frac{m}{4}[a^{2}+b^{2}-2c^{2}]$

$I_{zR}-I_{zS}=\frac{m}{4}[a^{2}+b^{2}-2ab]=\frac{m}{4}(a-b)^{2}\; \; \; \; (c^{2}=ab)$

$I_{zR}-I_{zS} >0$

$\frac{I_{zR}}{I_{zS}} >1$

Posted by

infoexpert23

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