Question

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

• A. $0.1$
• B. $0.7$
• C. $5$
• D. $10$

Answer 1

The correct option is B $0.7$

Here, the given body executes circular motion and hence, the centripetal force is supplied by the frictional force.

Equating the two forces, we have

$\mu mg=\frac{m{v}^2}{r}=mr{\omega }^2$
$\mu g=r{\omega }^2$
$0.5\times 9.8=10\times {\omega }^2$
$\omega =0.7rad{\sec }^{-1}$

The required angular velocity so that the body does not slip is $0.7rad{\sec }^{-1}$