Question

A coin just remains on a disc rotating at $120$ r.p.m. when kept at a distance of $1.5cm$ from the axis of rotation. Find the coefficient of friction between the coin and the disc.

• A. $0.965$
• B. $0.2414$
  
• C. $0.3231$
  
• D. $0.326$
  

Answer 1

The correct option is B $0.2414$

Given : $r=1.5cm=\frac{1.5}{100}m$

Frequency of rotation $\nu =120r.p.m=\frac{120}{60}rps=2rps$

Angular velocity of rotation of disc $\omega =$ $2\pi \nu$ $=$ $2\pi \times 2$ $=4\pi$ rad/sec

Equate the Centripetal force and force of friction, we get

$m{\omega }^{2}r=\mu mg$
Or ${\omega }^{2}r=\mu g$

Putting the values, we get

$\frac{(4\pi {)}^2\times 1.5}{100}$ $=\mu \times 9.8$

$⟹\mu =0.2414$

Hence Option (B) is correct .