Question

A coin is placed on a revolving disc which revolves at $60$ r.p.m. It does not slip-off when it is at $15cm$ from the axis of rotation. What should be the distance of the coin from the axis of rotation so that it does not slip-off, when the speed of the revolving disc is changed to $75$ r.p.m? (in $cm$)

• A. $9.6$
  
• B. $9.4$
  
• C. $8$
  
• D. $2$
  

Answer 1

The correct option is A $9.6$

at $x_1=15cm$ the coin doesn't slip, it means static friction force is balancing centrifugal force(assume mass of coin is $m$)

$\Rightarrow f_c=\mu mg=m{\omega }_1^2{x}_1...\left(1\right)$

if we change speed by $75rpm$, again friction force will balance out the centrifugal force

$\Rightarrow f_c=\mu mg=m{\omega }_2^2{x}_2...\left(2\right)$

by equation $\left(1\right)$ and $\left(2\right)$

$\Rightarrow x_2=\frac{{\omega }_1^2{x}_1}{{\omega }_2^2}=\frac{{\left(60\right)}^2\times 15}{{\left(75\right)}^2}=9.6cm$