Final answer:
The new angular velocity of the merry-go-round can be calculated by using the conservation of angular momentum, which remains constant when a child hops on. To find the new angular velocity, consider both the initial total angular momentum and the new increased moment of inertia due to the child hopping onto the merry-go-round.
Step-by-step explanation:
The question deals with the concept of conservation of angular momentum in physics. A merry-go-round with a moment of inertia of 319kg·m2 and an initial angular velocity of 2.33rad/s has a new angular velocity once a 25.8kg kid hops onto it at a distance of 1.40m from the axis. To solve for the new angular velocity, we must use the conservation of angular momentum, assuming no external torques act on the system.
The initial angular momentum Linitial is given by the product of the merry-go-round's moment of inertia I and its angular velocity ωinitial (Linitial = I × ωinitial). When the child hops on, the new total moment of inertia Itotal is the sum of the merry-go-round's moment of inertia and the child's moment of inertia (Itotal = I + mchild × r2). In this case, the child’s moment of inertia Ichild = mchild × r2 = 25.8kg × (1.40m)2.
Conservation of angular momentum implies Linitial = Lfinal, or I × ωinitial = Itotal × ωfinal, with ωfinal being the new angular velocity we want to find.
Therefore, to calculate the new angular velocity ωfinal, rearrange the equation to ωfinal = (I × ωinitial) / Itotal, and substitute the values in to solve for ωfinal.