A50.0 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.

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.

Ithink it is a. but i'm not 100 percent sure.