Question

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of $3.5$ revolutions per second. A coin placed at a distance of $1.25cm$ from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is $(g=10m/s^2)$

• A. $0.5$
• B. $0.7$
• C. $0.3$
• D. $0.6$

Answer 1

The correct option is D $0.6$

$3.5rev/second$

$1rev\to 2\pi rad$

$3.5rev\to 2\pi \times 3.5rad$

$\Rightarrow w=7\pi rad/sec$

$\mu mg=\frac{m{v}^2}{1.25}$

$\mu mg=\frac{m(rw{)}^2}{r}$

$\mu mg=mr{w}^2$

$\mu =\frac{r{w}^2}{g}=\frac{1.25\times {10}^{-2}\times {(7\times \frac{22}{7})}^2}{10}$

$=\frac{1.25\times {10}^{-2}\times {22}^2}{10}$

$=0.6$