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.
Work equals force × displacement (distance between initial point and end point is displacement) if u follow this it becomes work = 50 × 2 which is equal to 100
Frictional force always acts parallel to two planes in contact with each other and in a direction opposite to that of relative motion of the two bodies. 2. Frictional forces are caused due to intermolecular interactions between the bodies. Frictional force is more for rough surface and less for smooth surfaces.
