circular motion question

Question

A 67 gram coin sits on a horizontally rotating turntable. The turntable makes 1.7 revolution each second. The coin is located 10 cm from the axis of rotation of the turntable. The coin will slide off the turntable if it is located more than 14.6 cm from the axis of rotation. Calculate the coefficient of static friction? (Your result must contain 2 digits after the decimal and 1% of error will be tolerated in your answers. Take g=9.81 m/s2.)

Answer 1

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 = rω² = (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 s²) = 1.12 N

The formula for the force of static friction is:

F_f = µ_sF_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_sF_N = µ_smg = ma_c

µ_sg = a_c

µ_s = g ÷ a_c = (9.81 m s^-2) ÷ 16.66 m s^-2 = 0.59