Question

A block of mass $m$ is kept on a horizontal ruler. The friction coefficient between the ruler and the block is $\mu$. The ruler is fixed at one end and the block is at a distance $L$ from the fixed end. The ruler is rotated about the fixed end in the horizontal plane. What is the angular speed (maximum) for which the block does not slip ?

• A. $\sqrt{\frac{\mu g}{2L}}$
• B. $\sqrt{\frac{3\mu g}{2L}}$
• C. $\sqrt{\frac{2\mu g}{L}}$
• D. $\sqrt{\frac{\mu \text{g}}{L}}$
  

Answer 1

The correct option is D $\sqrt{\frac{\mu \text{g}}{L}}$



Taking the frame of reference fixed on the ruler,

From the FBD, the centrifugal force $\left(mr{\omega }^2\right)$ will be balanced by the friction force $f$.
Normal reaction $N$ will balance the weight of the block . So, the block of mass $m$ is in equllibrium.
In horizontal direction,
$f=mr{\omega }^2$ ........(1)
For maximum angular speed,
$f=f_{max}=\mu N$........(2)
In vertical direction,
$N=mg$ ........(3)

From (2) and (3)
$f=\mu mg$
Putting this in (1)
$\mu mg=mr{\omega }_{max}^2$ [where $r=L$ ]
$\Rightarrow {\omega }_{max}=$$\sqrt{\frac{\mu g}{L}}$