A playground merry-go-round has a radius 2.6 m and a rotational inertia 83 kg.m^2. When the merry-go-round is at rest, a child with mass 27 runs with speed 4.3 along a line tangent to the rim and jumps on. The angular velocity of the merry-go-round is then:

To find the angular velocity of the merry-go-round after the child jumps on, we can use the principle of conservation of angular momentum.

The angular momentum of an object rotating about a fixed axis is given by the product of its moment of inertia (rotational inertia) and its angular velocity. Mathematically, it can be expressed as:

L = Iω

Where:
L is the angular momentum
I is the moment of inertia
ω is the angular velocity

Before the child jumps onto the merry-go-round, its moment of inertia can be given as 83 kg·m^2, and its angular velocity is 0 since it is at rest.

After the child jumps on, the total angular momentum should be conserved. Let's denote the mass of the child as m and the angular velocity of the merry-go-round after the child jumps on as ω'.

We can write the equation as:

0 = (Img) × 0 + (I + mR^2) ω'

The first term is the angular momentum of the child before jumping on, which is 0 because the child is not rotating. The second term represents the angular momentum of the combined system of the merry-go-round and the child after the jump.

R is the radius of the merry-go-round, given as 2.6 m.

Now we can solve for ω':

(I + mR^2) ω' = 0

(83 kg·m^2 + m(27 kg)(2.6 m)^2) ω' = 0

Simplifying the expression:

(83 kg·m^2 + 1827.6 m^2·kg) ω' = 0

Now, divide both sides by (83 kg·m^2 + 1827.6 m^2·kg) to solve for ω':

ω' = 0 / (83 kg·m^2 + 1827.6 m^2·kg)

Since the denominator is nonzero, the angular velocity ω' is 0.

Therefore, the angular velocity of the merry-go-round after the child jumps on is 0.