A 60 g disk sits on a horizontally rotating turntable. The turntable makes exact

Question

A 60 g disk sits on a horizontally rotating turntable. The turntable makes exactly 4 revolutions each second. The disk is located 14 cm from the axis of rotation of the turntable. (a) What is the frictional force acting on the disk? (b) The disk will side off the turntable i it is lecated at a radius larger The disk will slide off the turntable if it is located at a radius larger than 20 cm from the axis of rotation. what is the coefficient of static friction? eBook o Show My Work opton e Another Version

Explanation / Answer

As there is no movement radially on the turntable friction (F) is providing the centripetal force ..
F = mR² .. .. (1rev = 2 radians, = 2 / 4 rad/s)

F = 60*10-3  * 0.14* (2 / 4)2

= 0.021 N

Coefficient of friction .. = Friction / force normal to surface (weight mg)

Again friction = centripetal force (at 0.20m radius)
F = mR² = 60*10-3 * 0.20 * (2 / 4)2    = 0.02961 N

= 0.02961 N /  60*10-3 *9.8

= 0.05

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