Question

A body moves along a circular path of radius $10\, m$ and the coefficient of friction is $0.5$. What should be its angular speed in rad/s if it is not to slip from the surface? ( $g=9.8\text{ }m{{s}^{-2}}$ )

• A. 5
• B. 10
• C. 0.1
• D. 0.7

Answer 1

Answer: (D)

0.7

Answer 2

The necessary centripetal force to the coin is provided by the frictional force.
For moving on circular path without slipping, centripetal force must equal frictional force. That is,
$\frac{m{{v}^{2}}}{r}=\mu mg$
Or $mr{{\omega }^{2}}=\mu mg$
$(\therefore v=r\omega )$
Or $r{{\omega }^{2}}=\mu g$
Given, $r = 10m, \mu = 0.5 \,g = 9.8 \,ms^{-2}$
$\therefore 10\omega^2 = 0.5 \times 9.8$
or $\omega = \sqrt{\frac{0.5\times 9.8}{10}}$
$= 0.7\,rad/s$