Final answer:
The problem involves the conservation of angular momentum, which dictates that the initial linear momentum of the running child is equal to the final angular momentum of the system after the child jumps onto the merry-go-round. By equating these and solving for the angular speed, we calculate the new angular velocity of the system.
Step-by-step explanation:
The child jumping onto a stationary merry-go-round exemplifies the principle of conservation of angular momentum, which states that the total angular momentum of a system remains constant if no external torques act on it. In this case, the only forces acting are internal, so angular momentum is conserved. The child's linear momentum before jumping onto the merry-go-round is m⋅v, where m is the child's mass and v is the tangential velocity. This linear momentum is converted into angular momentum when the child jumps onto the merry-go-round.
Initial linear momentum of the child is L = m⋅v⋅r, where r is the radius of the merry-go-round. The angular momentum must be the same before and after the child jumps on, so we equate initial angular momentum to the final angular momentum of the combined system (child plus merry-go-round). Using the moment of inertia of the child (I_child = m⋅r^2) and the merry-go-round (I_merry-go-round), the conservation of angular momentum is written as m⋅v⋅r = (I_merry-go-round + I_child)⋅ω, where ω is the angular speed after the child jumps on.
By rearranging this equation and solving for ω, we find the angular speed of the system after the child jumps on. Substituting all known values, we can calculate the desired angular speed.