A uniform stick of length `L` and mass `M` hinged at end is released from rest at an angle `theta_(0)` with the vertical show that when the angle with the vertical is `theta`. The hinge exerts of force `F_(r)` along the stick and `F_(t)` perpendicular tot he stick given by `F_(r)=(1)/(2)Mg(5costheta-3costheta_(0))` and `F_(t)=(1)/(4)Mgsintheta`

Explanation

(i) C is the centre of mass of the rod. Let `omega` be the angular speed of rod about point O at angle `theta`. From conservation of mechanical energy.
`Mg(L)/(2)(costheta-costheta_(0))=(1)/(2)((ML^(2))/(3))omega^(2)`
`thereforeomega^(2)=(3g)/(L)(costheta-costheta_(0))` ..(i)
Now, `F_(r)-Mgcostheta=M((L)/(2))omega^(2)` ..(ii)
From eqs. (i) and (ii) we get
`F_(r)=(1)/(2)Mg(5costheta-3costheta_(0))` hence proved
(ii) angular acceleration of rod at this instant
`alpha=(tau)/(I)=(Mg(L)/(2)sintheta)/((ML^(2))/(3))`
`=(3)/(2)(gsintheta)/(L)`
Tangential acceleration of COM.
`a_(t)=(alpha)((L)/(2))=(3)/(4)gsintheta` ...(iii)
Now `F_(t)+Mgsintheta=Ma_(t)` ....(iv)
from eqs. (iii) and (iv), we get
`F_(t)=-(1)/(4)Mgsintheta`
Here negative sign implies that direction of `F_(t)` is opposite to the component `Mgsintheta`.