Two identical masses on a horizontal turntable are located at different distances from the center of the turntable. As the turntable's rotation rate increases, which mass starts to slide off the turntable first?

1. The mass closest to the center.
2. The mass furthest from the center.
3. Both start to slide at the same time.

With the turntable spinning slowly, and both masses rotating with the turntable, what does the free-body diagram of each mass look like?

The only vertical forces are the normal force and the force of gravity, and these must balance. N = mg.

The only horizontal force is friction. If we turned friction off while the turntable was spinning the masses would go in a straight line, which would carry them away from the center of the turntable. Friction opposes this motion, so it points toward the center of the turntable. With the masses initially not slipping on the turntable (no relative motion) this is the static force of friction.

Apply Newton's second law in the x direction, with x horizontal and +x toward the center of the turntable:

SFx

=

max

=

m v2 r

fs

=

m v2 r

Note that as the speed of the masses increases the force of static friction increases. Once f_s reaches its maximum value the masses will begin to slide. The maximum f_s is the same for the two masses, but does one mass reach this maximum value before the other?

Each mass has a different speed, but the masses have the same angular velocity. Remember that v = w r. The equation for f_s can then be written as:

f_s = mw ²r

Because f_s is proportional to r, the force of static friction for the outer mass is larger than that for the inner mass. The outer mass will reach the maximum value of f_s first, so it will slide off the turntable first.