Question

A uniform slender bar AB of mass m and length L supported by a frictionless pivot at A is released from rest at its vertical position as shown in figure. Calculate the magnitude of the reaction at pivot when the bar just acquires the horizontal position shown dotted.

• A. $\frac{\sqrt{37}}{4}mg$
• B. $\frac{5}{4}mg$
• C. $\frac{\sqrt{63}}{4}mg$
• D. None of these

Answer 1

The correct option is A

$\frac{\sqrt{37}}{4}mg$

Since the pivot is frictionless and there are no other un-conservative forces, conservation of mechanical energy can be applied.

At this position (vertical)

$K_i+u_i=0+mg\left(\frac{\ell }{2}\right)$

At horizontal position

$k_f+u_f=\frac{1}{2}I{\omega }^2+0$

$k_i+u_i=k_f+u_f$[law of conservation of energy]

$mg\left(\frac{l}{2}\right)=\frac{1}{2}I{\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{mgl}{I}}=\sqrt{\frac{mgl}{\frac{\left(m{l}^2\right)}{3}}}=\sqrt{\frac{3g}{I}}[I=\frac{m{l}^2}{3}$

about [Remember this is $\omega$ at horizontal position] pivot, as rod is hinged at end]

Considering free body diagram of the rod in horizontal position.

The normal reaction due to hinge can be resolved as $N_{x}and{N}_y$ in x and y directions

τ on the rod = mg($\frac{\ell }{2})$

as τ = I$\alpha$

$mg\left(\frac{l}{2}\right)=\frac{m{l}^2}{3}\alpha$

$\alpha =\frac{3g}{2l}$

$N_x=m{a}_x[a_x$ is the centripetal acceleration i.e., towards the centre]

=$m.{\omega }^2\left(\frac{l}{2}\right)$

=$m\frac{3g}{l}.\frac{l}{2}=\frac{3mg}{2}$

$N_y=mg-m{a}_y$

$=mg-m.\left(\alpha \frac{l}{2}\right)$

$[a_y$ is the tangential acceleration = angular acceleration $\times$ 1 distance of com from axis]

$=mg-m\left(\frac{l}{2}\right)\left(\frac{3g}{2l}\right)$

$=mg-\frac{3mg}{4}=\frac{mg}{4}$

so the total normal reaction magnitude

$|\overrightarrow{N}|=\sqrt{N_x^2+N_y^2}$

$=\sqrt{(\frac{mg}{4}{)}^2+(\frac{3mg}{2}{)}^2}=\frac{\sqrt{37}}{4}mg$

if $\theta$ is the angle with the horizontal then

$tan\theta =\frac{N_y}{N_x}=\frac{1}{6}\Rightarrow \theta =ta{n}^{-1}\left(\frac{1}{6}\right)$