101. (a) The distance traveled by the coin in 3.14 s is 3(2πr) = 6(0.050) = 0.94 m. Thus, its speed is v= 0.94/3.14 = 0.30 m/s. (b) The centripetal acceleration is given by Eq. 6-17: 222(0.30 m/s)1.8 m/s .0.050 mvar===(c) The acceleration vector (at any instant) is horizontal and points from the coin towards the center of the turntable. (d) The only horizontal force acting on the coin is static friction fsand must be large enough to supply the acceleration of part (b) for the m= 0.0020 kg coin. Using Newton’s second law, ()230.0020 kg 1.8 m/s3.610Nsfma−=×.(e) The static friction
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.