Question

A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg-m2

and is rotating at 10 rev/min. A 25-kilogram child at rest jumps onto the edge of the merry-go-round. What is the new angular speed of the merry-go-round?

Answer 1

The question asks for the new angular speed of a merry-go-round after a child jumps onto it, using the principle of conservation of angular momentum in physics.

Step-by-step explanation:

The question involves applying the principle of conservation of angular momentum in rotational dynamics. A 25-kilogram child jumps onto a spinning merry-go-round with a radius of 2.0 meters and a moment of inertia of 250 kg-m2, which is initially rotating at 10 revolutions per minute. To find the new angular speed, we must consider the initial angular momentum of the system and the additional angular momentum contributed by the child when they jump onto the merry-go-round.

First, we need to convert the rotational speed to radians per second (the SI unit for angular velocity), and then calculate the initial angular momentum of the system. The child's linear velocity at the edge of the merry-go-round is tangential and will contribute to the angular momentum. By conserving angular momentum, the final angular momentum of the system (merry-go-round plus child) is equal to the initial angular momentum, which will allow us to solve for the final angular velocity.