A playground merry-go-round of radius R = 1.1 m has a moment of inertia of I = 280 kg*m^2. and is rotating at a rate of ω = 9 rev/min around a frictionless vertical axis. Facing the axle, a 37 kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round (in rev/min)?

15.2

5.987 To find the new angular speed of the merry-go-round, we can apply the law of conservation of angular momentum. According to this law, the initial angular momentum of the system should be equal to the final angular momentum.

The initial angular momentum of the system is given by:

L1 = I * ω1

Where:
- L1 represents the initial angular momentum of the system
- I is the moment of inertia of the merry-go-round
- ω1 is the initial angular speed of the merry-go-round

The final angular momentum of the system is given by:

L2 = (I + m * r^2) * ω2

Where:
- L2 represents the final angular momentum of the system
- m is the mass of the child
- r is the radius of the merry-go-round
- ω2 is the final angular speed of the merry-go-round

In this case, we have the values of R (radius of the merry-go-round), I (moment of inertia of the merry-go-round), ω1 (initial angular speed of the merry-go-round), m (mass of the child), and we need to find ω2 (final angular speed of the merry-go-round).

Substituting the values into the equations, we have:

L1 = I * ω1 = 280 kg*m^2 * (9 rev/min)

L2 = (I + m * r^2) * ω2 = (280 kg*m^2 + 37 kg * (1.1 m)^2) * ω2

Since angular momentum is conserved, we can set L1 equal to L2 and solve for ω2:

280 kg*m^2 * (9 rev/min) = (280 kg*m^2 + 37 kg * (1.1 m)^2) * ω2

Now we can solve for ω2 by rearranging the equation:

ω2 = (280 kg*m^2 * (9 rev/min)) / (280 kg*m^2 + 37 kg * (1.1 m)^2)

Simplifying and converting the answer from rad/s to rev/min, we obtain the new angular speed of the merry-go-round (in rev/min).