A playground merry-go-round of radius R = 2.15 m has a moment of inertia of I = 256 kg * m2 and is rotating at 10.9 revolutions/min about a frictionless vertical axle. Facing the axle, a 29.6 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed

ok, change angular speed from revs/min to radians/sec.

then the law of momentum takes place
Initial momentum=final momentum

I wi= (I + masskid*radius^2) wf
solve for wfinal wf.

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

The formula for angular momentum is L = I * ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.

Initially, the only object on the merry-go-round is the merry-go-round itself. So the initial angular momentum can be calculated as:

L_initial = I * ω_initial

Given:
Radius, R = 2.15 m
Moment of inertia, I = 256 kg * m^2
Initial angular speed, ω_initial = 10.9 revolutions/min

First, we need to convert the initial angular speed from revolutions/min to radians/s. Since 1 revolution is equal to 2π radians, we have:

ω_initial = 10.9 revolutions/min * (2π radians/1 revolution) * (1 min/60 s)

Let's calculate ω_initial:

ω_initial = 10.9 * (2π/60) radians/s

Now, we can substitute the values into the formula to find L_initial:

L_initial = I * ω_initial

Next, after the child hops onto the merry-go-round, the system consists of both the merry-go-round and the child. The total angular momentum of the system is conserved, so the new angular momentum is equal to the initial angular momentum.

The child's angular momentum can be calculated as:

L_child = m * r * v

Where m is the mass of the child, r is the radius of the merry-go-round, and v is the velocity of the child.

Given:
Mass of the child, m = 29.6 kg
Radius of the merry-go-round, r = R = 2.15 m

Now, we need to find the velocity of the child. Since the child hops onto the merry-go-round at the edge, the child's velocity will be equal to the tangential speed of the edge.

The tangential speed can be calculated as:

v = ω_final * r

Where ω_final is the final angular speed of the system.

Since the total angular momentum is conserved, we have:

L_initial = L_child

Substituting the values, we have:

I * ω_initial = m * r * v

Substituting the expression for v, we have:

I * ω_initial = m * r * (ω_final * r)

Now, we can solve for ω_final, the final angular speed:

ω_final = (I * ω_initial) / (m * r^2)

Substituting the values:

ω_final = (256 kg * m^2 * 10.9 * (2π/60) radians/s) / (29.6 kg * (2.15 m)^2)

Now, let's calculate ω_final to find the new angular speed of the merry-go-round!