An object is placed on a turntable (rotating platter) 25.0 cm from the center of rotation. The coefficient of friction between the turntable and object is 0.40. The turntable starts from rest (at time t = 0) with and is accelerated at 0.10 rad/s². The object will begin to slide off the turntable after A) 20 s. B) 40 s. C) 60 s. D) Not enough information is given to solve this problem.

Question

Quizplus · Accepted Answer

The object will begin to slide when the frictional force is no longer sufficient to provide the necessary centripetal force. The maximum static frictional force is μmg, and the required centripetal force is mω²r. Setting these equal gives μg = ω²r. Solving for ω using ω = αt, we find t = sqrt(μg/αr) = sqrt(0.40*9.8/(0.10*0.25)) = 40 s.

An Object Is Placed on a Turntable (Rotating Platter) 25