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Zepler [3.9K]
3 years ago
9

A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What

is the minimum coefficient of static friction between the turntable and the coin if the coin is not to slip?
Physics
1 answer:
Crank3 years ago
3 0

Answer:

\mu = \frac{r (2\pi f)^{2}}{g}

Explanation:

N = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

N = mg

f = frequency of rotation

Angular velocity of turntable is hence given as

w = 2\pi f

r = distance from the axis of rotation

\mu = minimum coefficient of static friction

static frictional force is given as

f = \mu N\\f = \mu mg

The  static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}

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