Question

A uniform rod of mass $m$ and length $l$ is released from rest in the vertical position on a rough (surface is sufficiently rough to prevent sliding) square corner A, shown in figure, then choose correct options: (given that normal acts along the length of the rod)

• A. If the rod just begins to slip when $\theta ={37}^{\circ }$ with vertical, then the coefficient of static friction $\mu$ between the rod and the corner is 0.3.
• B. If the end of the rod is notched so that it cannot slip, then the angle $\theta$ at which contact between the rod and the corner ceases is ${53}^{\circ }$
• C. If the rod just begins to slip when $\theta ={37}^{\circ }$ , then the coefficient of static friction $\mu$ between the rod and the corner is $\frac{3}{4}$
• D. If the end of the rod is notched so that it cannot slip, then the angle $\theta$ at which contact between the rod and the corner ceases is ${37}^{\circ }$.

Answer 1

Answer: (A), (B)

The correct options are
A If the rod just begins to slip when $\theta ={37}^{\circ }$ with vertical, then the coefficient of static friction $\mu$ between the rod and the corner is 0.3.
B If the end of the rod is notched so that it cannot slip, then the angle $\theta$ at which contact between the rod and the corner ceases is ${53}^{\circ }$


Using conservation of energy
$mg\frac{l}{2}=mg\left(\frac{l}{2}\right)\cos \theta +\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$
$\frac{{\omega }^{2}l}{2}=\frac{3g}{2}(1-\cos \theta )....\left(1\right)$
Using Newton's second law along the rod
$mg\cos \theta -N=m{\omega }^2\left(\frac{l}{2}\right)$
$N=mg\cos \theta -m\frac{3g}{2}(1-\cos \theta )......\left(2\right)$
$N=\frac{mg}{2}(5\cos \theta -3)$
Torque equation about point of contact
$\left(mg\sin \theta \right)\frac{l}{2}=\frac{m{l}^2}{3}\alpha$
$\alpha =\frac{3g\sin \theta }{2l}...\left(3\right)$
Using Newtons second law perpendicular to rod,
$mg\sin \theta -f_s=m\left(\frac{l}{2}\right)\alpha ....\left(4\right)$
$mg\sin \theta -f_s=m\left(\frac{l}{2}\right)\frac{3g\sin \theta }{2l}$
$f_s=\frac{mg\sin \theta }{4}....\left(4\right)$
When slipping begins
$f_s=\mu N.....\left(5\right)$
$\frac{mg\sin \theta }{4}=\mu \frac{mg}{2}(5\cos \theta -3)$
when $\theta ={37}^{\circ }$
$\mu =0.3$
For contact to cease $N=0\Rightarrow \cos \theta =\frac{3}{5}\Rightarrow \theta ={53}^{\circ }$