A 50.0 kg child stands at the rim of a merry-go-round of radius 1.50 m, rotating with an angular speed of 3.00 rad/s. (a) what i
s the child's centripetal acceleration? incorrect: your answer is incorrect. m/s2 (b) what is the minimum force between her feet and the floor of the merry-go-round that is required to keep her in the circular path? n (c) what minimum coefficient of static friction is required? is the answer you found reasonable? in other words, is she likely to be able to stay on the merry-go-round?
Weight of the child m = 50 kg Radius of the merry -go-around r = 1.50 m Angular speed w = 3.00 rad/s (a)Child's centripetal acceleration will be a = w^2 x r = 3^2 x 1.50 => a = 9 x 1.5 Centripetal Acceleration a = 13.5m/sec^2 (b)The minimum force between her feet and the floor in circular path Circular Path length C = 2 x 3.14 x 1.50 => c = 3 x 3.14 => C = 9.424 Time taken t = 2 x 3.14 / w => t = 6.28 / 3 => t = 2.09 Calculating velocity v = distance / time = 9.424 / 2.09 m/s => v = 4.5 m/s Calculating force, from equation F x r = mv^2 => F = mv^2 / r => 50 x (4.5)^2
/ 1.5 F = 50 x 3 x 4.5 => F = 150 x 4.5 => F = 675 N (c)Minimum coefficient of static friction u F = u x m x g => u = F / m x g => u = 675/ 50 x 9.81 => 1.376 u = 1.376 Hence with the force and the friction coefficient she is likely to stay on merry-go-around.
Carbon, is the group 14 element, with four electrons in its outer shell. Carbon always shares electrons to reach a complete valence shell, making bonds with other atoms.
