Brian L. answered • 11/25/19

Pharmacist specializing in many areas of math and science

Let ω = angular speed = 1.7 rev s^{-1 }= 3.4π rad s^{-1}

Let r = distance from the coin to the axis of rotation = 14.6 cm = 0.146 m

Let v_{t} = tangential speed = rω = (0.146 m)(3.4π rad s^{-1})

Let a_{c} = centripetal acceleration = rω^{2} = (0.146 m)(3.4π rad s^{-1})^{2} =

The coin will slide off the turntable when the centripetal force (Fc) exceeds the force of static friction (F_{f}).

The centripetal force is calculated as:

F_{c} = ma_{c} = (67 g)(0.146 m)(3.4π rad s^{-1})^{2}(10^{-3} kg g^{-1})(1 N kg^{-1} m^{-1} s^{2}) = 1.12 N

The formula for the force of static friction is:

F_{f} = µ_{s}F_{N}, where µ_{s }is the coefficient of static friction and F_{N} is the normal force.

The normal force is the force exerted on the coin by the surface of the turntable. It is equal in magnitude and opposite direction to the weight of the coin:

F_{N }= mg

We can now solve for the coefficient of static friction by setting F_{f }equal to F_{c}:

µ_{s}F_{N} = µ_{s}mg = ma_{c}

µ_{s}g = a_{c}

µ_{s} = g ÷ a_{c }= (9.81 m s^{-2}) ÷ 16.66 m s^{-2} = 0.59