Question

A uniform slender bar $AB$ of mass $m$ and length $L$ supported by a frictionless picot at $A$ is released from rest at its vertical position as shown in the figure. If the reaction at pivot when the bar just acquires the horizontal position shown dotted is $\frac{\sqrt{c}}{4}mg$, find the value of $c$.

Answer 1

As there is no friction, the mechanical energy will remain constant. Let us apply the conservation of mechanical energy between the vertical position $\left(A\right)$ and the horizontal position $\left(B\right)$.
Using $\Delta K+\Delta U=0$ $(\frac{1}{2}{I}_A{\omega }^2-0)+(-mg\frac{L}{2})=0$
which gives $\omega =\sqrt{\frac{3g}{L}}$
Considering the free-body diagram of the bar at this instant, the velocity of COM of bar at this instant is given below.
Let $\alpha$ be the angular acceleration of bar about $A$ at the instant when it becomes horizontal.
$\alpha =\frac{{\tau }_A}{I_A}=\frac{mg\frac{L}{2}}{\frac{m{L}^2}{3}}=\frac{3}{2}\frac{g}{L}$
Let $N_x$ and $N_y$ are the components of reaction at $A$. Then in horizontal direction $N_x=m{a}_r=m{\omega }^2\left(\frac{L}{2}\right)$ in vertical direction and $mg-N_y=m{a}_t........\left(i\right)$
and $a_t=$acceleration of COM in y-direction $=\left(L/2\right)a$
or, $N_y=mg-m\left(\frac{L}{2}\alpha \right)=mg-\frac{3mg}{4}=\frac{mg}{4}........\left(ii\right)$
Therefore, resultant reaction at support $A$ is $N=\sqrt{{N_x}^2+{N_y}^2}$
$N=\sqrt{{\left(\frac{3mg}{2}\right)}^2+{\left(\frac{mg}{4}\right)}^2}=\frac{\sqrt{37}}{4}mg$
$\tan \alpha =\frac{N_y}{N_x}=\frac{mg/4}{3mg/2}=\frac{1}{6}\Rightarrow \alpha ={\tan }^{-1}\left(\frac{1}{6}\right)$
Therefore, reaction at pivot is $\frac{\sqrt{37}}{4}mg$ at an angle
$\alpha ={\tan }^{-1}\left(\frac{1}{6}\right)$ with horizontal.