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):

  39.2 m/s

Explanation:

Given:

  free-fall from rest for 4 seconds

Unknown:

  final velocity (v)

Solution:

  equation: v = at

The acceleration due to gravity is 9.8 m/s², so the velocity after 4 seconds will be ...

  v = (9.8 m/s²)(4 s) = 39.2 m/s

At the bottom of the cliff, the velocity of the ball will be 39.2 m/s.