1a.A 0.500-kg stone is moving in a vertical circular path attached to a string t

Question

1a.A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0cm long. The stone is moving around the path at a constant frequency of 1.50 rev/sec. At the moment the stone is overhead,the stone is released. Calculate the velocity of the stone when it leaves the circular path.

1b.A 2000-kg car is traveling on a banked curve icy road. The road is banked at an angle of 12.0 degrees and has a radius of curvature of 500m. What is the velocity of the car necessary to travel on the icy road without sliding?

1c.A 2000-kg car is traveling on a banked curved icy road without sliding. The velocity of the car is 32.0m/s and the road is banked at an angle of 20.0 degrees. What is the radius of curvature of the road?

Explanation / Answer

1a. given mass of the stone m = 0.500 kg radius of the string r = 75cm                                  = 0.75 m frequency f = 1.50 rev/sec ....................................................... first convert the frequecy into angular velocity () rad/sec the relation between angular velocity and tangential velocity is v = r .......(1) -------------------------------------------------------------------------------- 1b. masss of the car m = 2000 kg radius of the curvature = 500 m angle of banking = 120 Banking angle of the roads tan = v2/rg   ...... (2) from the equation 2 we have v = sqrt [tan.rg]    = sqrt [tan120*500m*9.8m/sec2]    = 32.27 m/sec. ------------------------------------------------------------------------------------------ 1c. masss of the car m = 2000 kg angle of banking = 200 velocity of the car v = 32 m/sec Banking angle of the roads tan = v2/rg   ...... (3) from the equation 3 we have r = v2/tan*g    = (32)2/tan200*9.8m/sec2    = 287.6 m from the equation 2 we have v = sqrt [tan.rg]    = sqrt [tan120*500m*9.8m/sec2]    = 32.27 m/sec. ------------------------------------------------------------------------------------------ 1c. masss of the car m = 2000 kg angle of banking = 200 velocity of the car v = 32 m/sec velocity of the car v = 32 m/sec Banking angle of the roads tan = v2/rg   ...... (3) from the equation 3 we have r = v2/tan*g    = (32)2/tan200*9.8m/sec2    = 287.6 m    = 287.6 m

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