Aplayground merry-go-round of radius r = 1.80 m has a moment of inertia i = 255 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. facing the axle, a 24.0-kg child hops onto the merry-go-round and manages to sit down on the edge. what is the new angular speed of the merry-go-round?

7 rpm = 0.73 rad/s

Step-by-step explanation:

R = Radius of merry-go-round = 1.8 m

= Moment of inertia of merry-go-round = 255 kg m²

= Moment of inertia of child

ω = 9 rev/min

m = Mass of child = 24 kg

From the conservation of angular momentum

∴ New angular speed of the merry-go-round is 7 rpm =

answer: the answer is (b) amie should have multiplied

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

and the volume of the cylinder will be

therefore, we have

thus, ammie should have multiplied the volume of the sphere