Answer to Question #88543 in Mechanics | Relativity for Briana Berrios

Question

a coin sits on the edge of a turntable platform with a diameter of 40.0 cm. The turntable is the turned on and slowly accelerates till the coin slips off. If the coefficent of static friction between the coin and the turn table is .5, then what was the instantaneous frequency that the turntable was spinning at when the coin flew off the turntable?

Answer 1

According to Newton's second law, the coin starts slipping when the centripetal force becomes equal to the force of friction. In mathematical form the process described in the problem looks like

$$F_\text{friction}=F_\text{centripetal},$$

$$\mu mg=mv^2/r,$$

where

$$v^2/r=\omega^2r=(2\pi n)^2r,$$

thus

$$\mu g=\omega^2r,$$

and

$$\omega=\sqrt{\frac{\mu g}{r}}=\sqrt{\frac{0.5\cdot9.8}{0.4}}=3.5\text{ rad/s},$$

or

$$n=\frac{\omega}{2\pi}=\frac{3.5}{2\cdot3.14}=0.56\text{ s}^{-1}.$$

The turntable did 0.56 revolution per second.when the coin flew off.