A merry-go-round can be modeled as a disk with radius 2.5 meters and mass of 150 kg. For this problem, assume the density is uniform. It is initially rotating at 2.0 rad/sec. Two children of mass 30 kg and 40 kg hop onto the merry-go-round and stay near the edge. What is the new angular speed?

angular momentum is conserved

intial =final
I *wi=(I *wf+70*r^2 *wf)

where I= 1/2*150
solve for wf

To find the new angular speed, we can apply the principle of conservation of angular momentum. This principle states that the total angular momentum of a system remains constant unless acted upon by an external torque.

The initial angular momentum of the merry-go-round is given by the formula:

L_initial = I * ω_initial

Where L_initial is the initial angular momentum, I is the moment of inertia, and ω_initial is the initial angular speed.

The moment of inertia of a disk can be calculated using the formula:

I = (1/2) * m * r^2

Where m is the mass of the disk and r is the radius.

In this case, the mass of the merry-go-round is 150 kg and the radius is 2.5 meters. Substituting these values into the formula, we get:

I = (1/2) * 150 kg * (2.5 meters)^2 = 468.75 kg*m^2

The initial angular momentum of the merry-go-round is then:

L_initial = 468.75 kg*m^2 * 2.0 rad/sec = 937.5 kg*m^2/s

When the children hop onto the merry-go-round, the moment of inertia of the system increases due to the additional mass located at the edge of the disk. The new moment of inertia can be calculated as:

I_new = I + (m1 * r^2) + (m2 * r^2)

Where m1 and m2 are the masses of the children and r is the radius of the merry-go-round.

Substituting the given values, we get:

I_new = 468.75 kg*m^2 + (30 kg * (2.5 meters)^2) + (40 kg * (2.5 meters)^2) = 1406.25 kg*m^2

Finally, using the principle of conservation of angular momentum, we can calculate the new angular speed, denoted as ω_new:

L_initial = L_new

I * ω_initial = I_new * ω_new

Solving for ω_new, we get:

ω_new = (I * ω_initial) / I_new

Substituting the known values, we have:

ω_new = (468.75 kg*m^2 * 2.0 rad/sec) / 1406.25 kg*m^2

Simplifying the expression, we get:

ω_new ≈ 0.667 rad/sec

Therefore, the new angular speed is approximately 0.667 rad/sec.