A playground merry-go-round (which can be modeled as a disk) has a 3.4 m diameter, a mass of 240 kg and is rotating at 22.5 rpm. A 35 kg child running at 6.0 m/s, tangent to and in the same direction as the merry-go-round is turning, jumps on the outer edge. Ignoring any friction, what is the angular velocity in rpm after the child jumps on?

To solve this problem, we can apply the principle of conservation of angular momentum. The angular momentum before the child jumps on is equal to the angular momentum after the child jumps on.

The formula for angular momentum is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Before the child jumps on, the only object with angular momentum is the merry-go-round. The moment of inertia of a disk is given by I = (1/2)mr^2, where m is the mass of the merry-go-round and r is the radius (which is half the diameter).

Let's calculate the initial angular momentum:

Moment of inertia (I) = (1/2) * mass (m) * radius^2
= (1/2) * 240 kg * (1.7 m)^2
= 163.2 kg*m^2

To convert the rotational velocity of the merry-go-round from rpm to rad/s, we can use the conversion factor: 1 rpm = (2π/60) rad/s.

Initial angular velocity (ω_initial) = 22.5 rpm * (2π/60) rad/s
= 7.48 rad/s

Now, when the child jumps on, the moment of inertia of the system (merry-go-round + child) changes. The new total moment of inertia (I_total) will be equal to the sum of the moment of inertia of the merry-go-round (I_initial) and the moment of inertia of the child (I_child).

The moment of inertia of a point mass rotating at a given radius is given by I = mr^2, where m is the mass of the object and r is the distance from the axis of rotation.

Since the child jumps on the outer edge, the distance from the axis of rotation will be equal to the radius of the merry-go-round (1.7 m).

The moment of inertia of the child (I_child) = mass (m_child) * radius^2
= 35 kg * (1.7 m)^2
= 102.9 kg*m^2

The new total moment of inertia (I_total) = I_initial + I_child
= 163.2 kg*m^2 + 102.9 kg*m^2
= 266.1 kg*m^2

Now, we can use the conservation of angular momentum to find the final angular velocity (ω_final):

L_initial = L_final

I_initial * ω_initial = I_total * ω_final

Solving for ω_final:

ω_final = (I_initial * ω_initial) / I_total

Plugging in the values:

ω_final = (163.2 kg*m^2 * 7.48 rad/s) / 266.1 kg*m^2
≈ 4.60 rad/s

To convert the final angular velocity from rad/s to rpm, we can use the conversion factor in reverse: 1 rad/s = (60/2π) rpm.

Final angular velocity in rpm ≈ 4.60 rad/s * (60/2π) rpm
≈ 27.8 rpm (rounded to one decimal place)

Therefore, the angular velocity in rpm after the child jumps on is approximately 27.8 rpm.