# Question #b80f7

Dec 24, 2016

The maximum speed the coin could reach without slipping would be $1.78 \frac{m}{s}$

#### Explanation:

Since the coin is travelling in a circle at constant speed, it requires a centripetal force

${F}_{c} = \frac{m {v}^{2}}{r}$

In this situation, it is the friction between the coin and the turntable that will provide this force

${F}_{f} = \mu m g$

Since the left side of these equations are equal (based on the previous sentence), the right sides must be equal as well

$\frac{m {v}^{2}}{r} = \mu m g$

Note that we are able to cancel the mass from each side of this equation. This tells us that any coin placed at this location would behave the same as this $25 g$ coin. The mass does not affect the outcome of the problem.
So,

${v}^{2} / r = \mu g$

and therefore

$v = \sqrt{\mu g r} = \sqrt{\left(1.625\right) \left(9.8\right) \left(0.2\right)} = 1.78 \frac{m}{s}$