Correct Answer - (a) `sqrt(3) m l omega^(2)`
(b) `(F_(net))_(x) = -(F)/(4), (F_(net))_(y) = sqrt(3) ml omega^(2)`.
The distance of centre of mass `(CM)` of the system about point `A` will be
`r = (l)/(sqrt(2))`
therefore, the magnitude of horizontal force exerted by the hinge on the body is cejntripetal force `F = (3m)r omega^(2)`
`F = (3m) ((l)/(sqrt(3))) omega^(2) rArr F - sqrt(3) ml omega^(2)`
(b) Angular acceleration of system about point `A` is
`alpha = (tau_(A))/(I_(A)) = ((F)(sqrt(3)/(2)l))/(2 ml^(2)) rArr alpha = (sqrt(3) F)/(4 ml)`
acceleration of `C.M` along x-axis
`a_(x) = r alpha rArr a_(x) = ((l)/(sqrt(3))) xx ((sqrt(3) F)/(4 ml)) rArr a_(x) = (F)/(4 m)`
let `F_(x)`, be the force applied by teh hinge along x-axis. Then,
`F_(x) + F = (3m) a_(x) rArr F_(x) + F = (3m) ((F)/(4 m))`
`F_(x) + F = (3)/(4) F` or `F_(x) = (F)/(4)`
let `F_(y)` be the force applied by the hinge along y-axis. Then,
`F_(y)` = centripetal froce `rArr F_(y) = sqrt(3) m l omega^(2)`.