Question

A coin is placed 10.8 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 51.1 rpm is reached, at which point the coin slides off. What is the coefficient of static friction μs between the coin and the turntable?

Answer 1

Coefficient of static friction = 0.547

Step-by-step explanation:

The maximum resistive force offered by the body(coin)against the applied force(turntable) to continue its state of motion is called Coefficient Of Static Friction

When given

The angular velocity which is 51.1rpm

We are going to calculate velocity of coin

= r*w(angular velocity) = 0.108*(51.1*2π/60) = 0.58

Coefficient=F/N

F= coefficient* N = coefficient*m*g

F= mv^2/r which is centripetal force

Therefore, mv^2/r = coefficient*m*g

coefficient = mv^2/mgr = v^2/rg

Where g,gravity = 9.81

Substitute into the equation coefficient = 0.58^2/9.81*0.108

= 0.547

Answer 2

Coefficient of static friction = 0.547

Step-by-step explanation:

The maximum resistive force offered by the body(coin)against the applied force(turntable) to continue its state of motion is called Coefficient Of Static Friction

When given

The angular velocity which is 51.1rpm

We are going to calculate velocity of coin

= r*w(angular velocity) = 0.108*(51.1*2π/60) = 0.58

Coefficient=F/N

F= coefficient* N = coefficient*m*g

F= mv^2/r which is centripetal force

Therefore, mv^2/r = coefficient*m*g

coefficient = mv^2/mgr = v^2/rg

Where g,gravity = 9.81

Substitute into the equation coefficient = 0.58^2/9.81*0.108

= 0.547