physics circular motion

Question

A 99 gram coin sits on a horizontally rotating turntable. The turntable makes 1.2 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 cm from the axis of rotation. Calculate the coefficient of static friction?Take g=9.81

Answer 1

We can set the force that the turntable exerts on the coin equal to the force of friction holding the coin on the table. In this condition, the coin is 14 cm from the axis of rotation:

F_TURNTABLE = F_FR

The force of friction is equal to the normal force multiplied by the static coefficient of friction. The normal force is equal to the gravitational force on the coin (m*g):

F_FR = μF_N = μmg

The force that the turntable exerts on the coin:

F_TURNTABLE = mrω²

Using the first equation, we now have:

μmg = mrω²

μg = rω² (masses cancel)

ω is angular velocity in radians per second, so we need to convert 1.2 revolutions per second:

1.2 rev/s * (2π rad/1 rev) = 2.4π rad/s

We can now solve for the coefficient of friction.

μ*(9.81 m/s²) = (0.14 m)*(2.4π rad/s)²

μ = 0.81