A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 260 kg · m2 and is rotating at 12.0 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?

The new angular speed of the merry-go-round is 8.31 rev/min.

Explanation:

Because the merry-go-round is rotating about a frictionless axis there’re not external torques if we consider the system merry-go-round and child. Due that we can apply conservation fo angular momentum that states initial angular momentum (Li) should be equal final angular momentum (Lf):

(1)

The initial angular momentum is just the angular momentum of the merry-go-round (Lmi) that because it's a rigid body is defined as:

(2)

with I the moment of inertia and ωi the initial angular speed of the merry-go-round

The final angular momentum is the sum of the final angular momentum of the merry-go-round plus the final angular momentum of the child (Lcf):

(3)

The angular momentum of the child should be modeled as the angular momentum of a punctual particle moving around an axis of rotation, this is:

(4)

with m the mass of the child, R the distance from the axis of rotation and vf is final tangential speed, tangential speed is:

(5)

(note that the angular speed is the same as the merry-go-round)

using (5) on (4), and (4) on (3):

(6)

By (5) and (2) on (1):

Solving for ωf (12.0 rev/min = 1.26 rad/s):

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west