A cube of side `a` and mass `M` is pivoted on a horizontal axis such that it can rotate about one of tis edges. Initially it is held in the highest position and then released. Find horizontal and vertical components of reaction of pivot on the cube at the moment when the reaction toward the centre of the cube becomes zero. Given that `M.I` of the cube about its edge is equal to `(2)/(3)Ma^(2)`.

Explanation

Let `P` and `Q` be the components of reaction of pivot on the cube along and perpendicular to the `OC`.
`M.I.` of the cube about `O` is `(2)/(3)Ma^(2)`
`alpha=(tau)/(I)=Mg sintheta.(a)/(sqrt(2))/(2)/(3)Ma^(2)=(3)/(2sqrt(2))(gsintheta)/(a)` Now along `OC`, `Mg costheta-P=M(a)/(sqrt(2))omega^(2)`
Perpendicular to `OC`, `Mgsintheta-Q=M(a)/(sqrt(2))alpha`
`Q=(Mgsintheta)//4`
Intergrating `alpha`, we get `omega^(2)=(3g(1-costheta))/(asqrt(2))`.......(`1`)
`P=Mg costheta-Ma(3g(1-costheta))/(2a)=(-3)/(2)Mg+(5)/(2)Mg costheta`
When `P=0` `costheta=3//5`, `sintheta=4//5`
Now horizontal component of reaction is `P sin theta-Q cos theta = (-3//25)Mg`
Vertical component of reaction is `P cos theta + Q sin theta = (4//25)Mg`