Question
A coin is placed 26 CM from the center of a horizontal turnable intially at rest. The turnable begin to ratate. When the speed of coin is 120 CMS rotating at constant rate the coin just begin to slip the acceleration of gravity is 980 CM/S2. What is the coefficient static friction between the coin and turnable?
Answers
R_scott
normal force = m g
centripetal force = m v^2 / r
μ static = centripetal / normal = v^2 / (g r)
centripetal force = m v^2 / r
μ static = centripetal / normal = v^2 / (g r)