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 is 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.