Question

Two children (m = 29.0 kg each) stand opposite each other on the edge of a merry-go-round. the merry-go-round, which has a mass of 1.68x10² kg and a radius of 1.4 m, is spinning at a constant rate of 0.48 rev/s. treat the two children and the merry-go-round as a system. (a) calculate the angular momentum of the system, treating each child as a particle. (give the magnitude.) kg · m²/s

Answer 1

The angular momentum of the system is 212.70 kg · m²/s.

Step-by-step explanation:

To calculate the angular momentum of the system, we need to consider the angular momentum of each child separately.

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

For each child, the moment of inertia is given by the formula I = mR^2, where m is the mass and R is the radius of the merry-go-round.

Using the values provided, the moment of inertia for each child is:

• I1 = mR^2 = (29.0 kg)(1.4 m)^2 = 57.68 kg · m²
• I2 = mR^2 = (29.0 kg)(1.4 m)^2 = 57.68 kg · m²

The total moment of inertia for the system is the sum of the individual moment of inertia:

I_total = I1 + I2 + I_merry-go-round = 57.68 kg · m² + 57.68 kg · m² + (1.68x10² kg)(1.4 m)^2 = 442.68 kg · m²

Now, we can calculate the angular momentum:

L = I_total * ω = 442.68 kg · m² * 0.48 rev/s = 212.70 kg · m²/s