Question

A small collar of mass $m=100\text{g}$ slides over the surface of a horizontal circular rod of radius $R=0.3\text{m}$. The coefficient of friction between the rod and the collar is $\mu =0.8$. When the speed of the collar is $V=2{\text{ms}}^{-1}$, the resultant force on rod is

• A. $\frac{\sqrt{21}}{3}N$
• B. $\frac{\sqrt{31}}{3}N$
• C. $\frac{\sqrt{41}}{3}N$
• D. $\frac{\sqrt{51}}{3}N$

Answer 1

The correct option is C $\frac{\sqrt{41}}{3}N$

For reference, FBD of collar




Vertical normal reaction ,
$N_V=mg=0.1\times 10=1\text{N}$

Horizontal normal reaction ,
$N_H=\frac{m{V}^2}{R}=\frac{0.1\times 2^2}{0.3}=\frac{4}{3}\text{N}$
Resultant normal reaction , $N=\sqrt{N_V^2+N_H^2}=\frac{5}{3}N$
Friction, $f=\mu N=\frac{4}{3}N$
Resultant force on rod, $R=\sqrt{N^2+f^2}=\frac{\sqrt{41}}{3}N$