Answer to Question #91706 in Classical Mechanics for Louie

Question

Question 19
A coin placed 0.4m from the centre of a rotating, horizontal turntable slips when its speed is 0.6 m/s. What is the coefficient of static friction between the coin and turntable?

Answer 1

At the moment when the coin starts to slide, centripetal force equals to the friction force:

$$F_{centr}=F_{fric }$$

That means:

$$\frac{mv^2}{r}=\mu mg$$

$$\mu=\frac{v^2}{rg}=\frac{0.6^2}{0.4*9.8}=0.092$$

So, the coefficient of friction is 0.092