Question
Physics Question on System of Particles & Rotational Motion
For the given uniform square lamina ABCD, whose centre is O,
A
2IAC=IEF
B
IAD=3IEF
C
IAD=4IEF
D
IAC=2IEF
Answer
IAD=4IEF
Explanation
Solution
Let the each side of square lamina is d. So, \hspace5mm I_{EF}=I_{GH} \hspace10mm (due \, to \, symmetry) and \hspace5mm I_{AC}=I_{BD} \hspace10mm (due \, to \, symmetry) Now, according to theorem of perpendicular axis \hspace10mm I_{AC}+I_{BD}=I_0 or \hspace10mm 2I_{AC}=I_0 \hspace5mm ...(i) and \hspace10mm I_{EF}+I_{GH}=I_0 or \hspace10mm 2I_{EF}=I_0 \hspace5mm ...(ii) From Eqs. (i) and (ii), we get IAC=IEF \therefore \hspace5mm I_{AD}=I_{EF}+ \frac {md^2}{4} \hspace5mm =\frac {md^2}{12}+ \frac {md^2}{4} \hspace5mm \bigg (as \, I_{EF}= \frac {md^2}{12}\bigg ) So, \hspace5mm I_{AD}= \frac {md^2}{3}=4I_{EF}