A small coin is placed at a distance of 1 m from the center of a horizontal turntable. The turntable, initially at rest, begins to rotate with a constant angular acceleration of 3 rad/s ^2 . If the coefficient of static friction between the coin and the turntable is 0.5 , at what time will the coin just begin to slip? (Take g = 10 m/s ^2 )

Step-by-step solution

The coin undergoes non-uniform circular motion, so it experiences both tangential and radial (centripetal) accelerations. The tangential acceleration a_t is constant: a_t = R = (3)(1) = 3 m/s ^2 The angular velocity at time t is = t = 3t . The radial acceleration a_r at time t is: a_r = ^2 R = (3t)^2(1) = 9t^2 m/s ^2 The net acceleration of the coin is the vector sum of these two perpendicular components: a_ net = a_t^2 + a_r^2 = 3^2 + (9t^2)^2 = 9 + 81t^4 The force required to provide this acceleration is m a_ net