A thin plank of mass `M` and length `l` is pivoted at one end. The plank is releaed at `60^(@)` from the vertical. What is the magnitude and direction of the force on the pivot when the plank is horizontal?

Explanation

Let `omega` be the angular velocity and `alpha` the angular acceleration of rod in horizontal position. Then
`alpha=((Mg)(l)/(2))/((Ml^(2))/(3))=(3)/(2)(g)/(l)` .(i)
`(1)/(2)((Ml^(2))/(3))omega^(2)=Mg(l)/(4)`
`thereforeomega=(3)/(2).(g)/(l)` ..(ii)
`F_(x)=M((l)/(2))omega^(2)=M((l)/(2))((3g)/(2l))=(3)/(4)Mg`
`Mg-F_(y)=M(alpha)=((l)/(2))` ltbr `F_(y)=Mg-(3)/(4)Mg=(Mg)/(4)`
`thereforeF=sqrt(F_(x)^(2)+F_(y)^(2))`
`=(sqrt(10))/(4)Mg`
`tanalpha=(F_(y))/(F_(x))=((Mg//4))/(3Mg//4))=(1)/(3)`
`thereforealpha=tan^(-1)((1)/(3))`