In a playground there is a small merry-go-round of radius 1.40 m and mass 140 kg. The radius of gyration is 0.960 m. A child of mass 43.0 kg runs at a speed of 3.70 m/s tangent to the rim of the merry-go-round while it is at rest, and then jumps on. Find the angular velocity (in radians/second) of the merry-go-round and child together. Neglect friction.

You must be having an identity crisis. You've posted physics questions under four different names in the last few minutes.

To find the angular velocity of the merry-go-round and the child together, we can apply the law of conservation of angular momentum, which states that the total angular momentum of a system remains constant if no external torques act on it.

The initial angular momentum of the merry-go-round is zero since it is at rest. The angular momentum of the child running tangent to the rim of the merry-go-round can be calculated using the formula:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

The moment of inertia of a point mass rotating about an axis at a distance r is given by:

I = m * r^2

where m is the mass and r is the distance from the axis of rotation.

First, let's calculate the initial angular momentum of the child:

I_child = m_child * r^2
= 43.0 kg * (1.40 m)^2
= 86.38 kg·m^2

Since the child is running tangent to the rim of the merry-go-round, the distance from the axis of rotation is the radius of the merry-go-round, which is 1.40 m.

Next, let's calculate the final moment of inertia of the child and the merry-go-round together. The moment of inertia for a compound system like this can be calculated using the parallel axis theorem:

I_total = I_merry-go-round + I_child

The parallel axis theorem states that the moment of inertia of a body about an axis parallel to an axis through its center of mass is given by:

I = I_cm + M * d^2

where I_cm is the moment of inertia about the center of mass, M is the total mass of the object, and d is the distance between the center of mass and the parallel axis of rotation.

The radius of gyration is defined as the distance from the center of mass to the parallel axis of rotation. Therefore, we can rewrite the parallel axis theorem as:

I = I_cm + m * k^2

where k is the radius of gyration.

Rearranging the equation, we can solve for I_cm:

I_cm = I - m * k^2

Plugging in the given values:

I_merry-go-round = m_merry-go-round * r^2
= 140 kg * (1.40 m)^2
= 274.40 kg·m^2

I_cm = I_merry-go-round - m_merry-go-round * k^2
= 274.40 kg·m^2 - 140 kg * (0.960 m)^2
= 274.40 kg·m^2 - 127.104 kg·m^2
= 147.296 kg·m^2

Finally, we can calculate the final angular momentum of the system:

L_total = I_total * ω_total

Since the initial angular momentum is zero and the final angular momentum is the sum of the angular momentum of the child and the merry-go-round, we have:

L_total = L_child + L_merry-go-round
= I_child * ω_child + I_merry-go-round * ω_merry-go-round

Since the child jumps onto the merry-go-round, their angular velocities become equal. So we can rewrite the equation as:

L_total = (I_child + I_merry-go-round) * ω_total

Now, solving for ω_total:

ω_total = L_total / (I_child + I_merry-go-round)

Let's plug in the values:

ω_total = 0 / (86.38 kg·m^2 + 274.40 kg·m^2)
= 0 radians/second

Therefore, the angular velocity of the merry-go-round and child together is 0 radians/second.