A square plate hinged at `A`, of side a and mass `M` is placed in `(x-z)` plane. The plate is allowd to fall upto `(x-y)` plane. Find its angular velocity

Options

A. `[(3g)/(asqrt(2))]^(1//2)`
B. `[(24g)/(7a)]^(1//2)`
C. `[(12 g)/(7a)]^(1//2)`
D. `[(12 g)/(7sqrt(2)a)]^(1//2)`

Answer

A

Explanation

`AC=asqrt(2)`
The plates falls by `AO=a//sqrt(2)`
`I`(diagonal `DB`)=`(I_(z))/2=(I_(1)+I_(2))/2`

`=1/2[(Ma^(2))/2]=(Ma^(2))/6`
Since `I_(1)=I_(2)=(Ma^(2))/6,I_(A)=I_(DB)+M(AO)^(2)`
`=(Ma^(2))/6+M(a//sqrt(2))^(2)=(2/3)Ma^(2)=2/3Ma^(2)`
`Mga/(sqrt(2))=1/2Iomega^(2)=1/2(2/3)Ma^(2)omega^(2)`
Which gives `omega=[(3g)/(2a)]^(1//2)=[(3a)/(asqrt(2))]^(1//2)`