Question

A uniform stick of length $L$ and mass $M$ hinged at one 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 a force $F_r$ along the stick and $F_t$ perpendicular to the stick given by
$F_r=\frac{1}{2}Mg(5\cos \theta -3\cos {\theta }_0)$ and $F_t=\frac{1}{4}Mg\sin \theta$

Answer 1

(i) $C$ is the centre of mass of the rod. Let $\omega$ the angular speed of rod about point $O$ at angle $\theta$ From
conservation
of mechanical energy,$Mg\frac{L}{2}(\cos \theta -\cos {\theta }_0)=\frac{1}{2}\left(\frac{M{L}^2}{3}\right){\omega }^2$
$\therefore {\omega }^2=\frac{3g}{L}(\cos \theta -\cos {\theta }_0)....\left(i\right)$
Now, $F_r-Mg\cos \theta =M\left(\frac{L}{2}\right){\omega }^2....\left(ii\right)$
From Eqs. (i) and (ii), we get
$F_r=\frac{1}{2}Mg(5\cos \theta -3\cos {\theta }_0)$ Hence proved.
(ii) Angular acceleration of rod at this instant,
$\alpha =\frac{\tau }{I}=\frac{Mg\frac{L}{2}\sin \theta }{\frac{M{L}^2}{3}}$
$=\frac{3}{2}\frac{g\sin \theta }{L}$
Tangential acceleration of COM,
$a_t=\left(\alpha \right)\left(\frac{L}{2}\right)=\frac{3}{4}g\sin \theta$.....(iii)
Now, $F_t+Mg\sin \theta =M{a}_t....\left(iv\right)$
From Eqs. (iii) and (iv), we get
$F_t=-\frac{1}{4}Mg\sin \theta$
Hence proved.
Here negative sign implies that direction of $F_t$ is opposite to the component $Mg\sin \theta$