Question

A rod is released from the position as shown in figure. If there is sufficient friction between the rod and the plane to prevent slipping, the minimum value of friction co-efficient between rod and ground is

• A. $0.6$
• B. $0.4$
• C. $0.5$
• D. $0.3$

Answer 1

The correct option is A $0.6$

If the lower end will not slip then rod will rotate about it, due to torque by gravitational force at the instant when it is released


Applying torque equation about the lower end, we have
$\tau =I\alpha$

$\Rightarrow mg\frac{L}{2}\cos {45}^{\circ }=\frac{m{L}^2}{3}\alpha$

Where, $\tau$ is the torque produced by the weight about lower end of the rod, $I$ is moment of inertia about lower end and $\alpha$ is angular acceleration.

$\Rightarrow \frac{3g}{2\sqrt{2}L}=\alpha ......\left(1\right)$

For centre of rod $\text{C}$,
Linear acceleration, $a_c=\frac{L}{2}\alpha$

From equation $\left(1\right)$

$\Rightarrow a_c=\frac{L}{2}\times \frac{3g}{2\sqrt{2}L}$

$\Rightarrow a_c=\frac{3g}{4\sqrt{2}}......\left(2\right)$

The horizontal component of $a_c$ will be due to frictional force at the lower end.
Balancing horizontal force,

$\therefore \Rightarrow \mu N=m{a}_c\cos {45}^{\circ }$

From $\left(2\right)$,

$\Rightarrow \mu N=\frac{3mg}{8}.....\left(3\right)$

For vertical motion, balancing force

$mg-N=m{a}_c\sin {45}^{\circ }$

From $\left(3\right)$,
$\Rightarrow N=mg-\frac{3mg}{8}.....\left(4\right)$

From $\left(3\right)$ and $\left(4\right)$,

$\mu (mg-\frac{3mg}{8})=\frac{3mg}{8}$

$\Rightarrow \mu \times \frac{5mg}{8}=\frac{3mg}{8}$

$\therefore \mu =0.6$

Hence, option $\left(a\right)$ is a correct option.