Question

A uniform rod of mass $m$ and length $l$ is released from rest in the vertical position ( $\theta =0^{\circ }$ ) on a rough surface (surface is sufficiently rough to prevent sliding) from a corner 'A'. Instataneous position of rod with $\theta$ is shown in figure. Untill rod does not leave contact with surface, then the correct options are

• A. Angular velocity of rod at any given instant will be given by $\sqrt{\frac{3g(1-\cos (\theta \left)\right)}{l}}$
• B. Angular acceleration of rod at any given instant will be given by $\frac{3g\sin \left(\theta \right)}{2l}$
• C. Normal reaction at any instant will be given by $mg\left(2\cos \left(\theta \right)-\frac{3}{2}\right)$
• D. At limiting value of static friction, coefficient of friction $\mu =\frac{\sin \left(\theta \right)}{2\left(5\cos \left(\theta \right)-3\right)}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A Angular velocity of rod at any given instant will be given by $\sqrt{\frac{3g(1-\cos (\theta \left)\right)}{l}}$
B Angular acceleration of rod at any given instant will be given by $\frac{3g\sin \left(\theta \right)}{2l}$
D At limiting value of static friction, coefficient of friction $\mu =\frac{\sin \left(\theta \right)}{2\left(5\cos \left(\theta \right)-3\right)}$


Initially, the rod was vertical and potential energy of rod by considering horizontal plane containing point A as zero potential energy reference level is $mg\frac{l}{2}$. As friction is static, so it can't do any work. So, by conservation of energy
$mg\frac{l}{2}=mg\left(\frac{l}{2}\right)cos\theta +\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2\dots \dots \left(1\right)$
Therefore, angular velocity of rod at any given instant will be given by $\sqrt{\frac{3g(1-\cos \theta )}{l}}$

And by writing torque equation about Point A,
$\left(mgsin\theta \right)\frac{l}{2}=\frac{m{l}^2}{3}\alpha \dots \dots \left(2\right)$
Therefore, angular acceleration of rod at any given instant will be given by $\frac{3g\sin \left(\theta \right)}{2l}$

Now, by writing equation of forces along direction of rod , we get,
$mgcos\theta -N=m{\omega }^2\left(\frac{l}{2}\right)\dots \dots \left(3\right)$
Normal reaction at any instant will be given by $mg\left(\frac{5\cos \left(\theta \right)-3}{2}\right)$

Now, by writing equation of forces perpendicular to rod
$mgsin\theta -f_s=m\left(\frac{l}{2}\right)\alpha \dots \dots \left(4\right)$
$\Rightarrow f_s=\frac{mg\sin \left(\theta \right)}{4}$
Also, at limiting value of static friction,
$f_s=\mu N\dots \dots \left(5\right)$
$\mu =\frac{\sin \left(\theta \right)}{2\left(5\cos \left(\theta \right)-3\right)}$