A playground merry-go-round of radius 1.92 m has a moment of inertia 411 kg m2 and is rotating at 10.4 rev/min. A child with mass 38.3 kg jumps on the edge of the merry-go-round. Assuming that the boy's initial speed is negligible, what is the new angular speed of the merry-go-round? Answer in units of rev/min.

To solve this problem, we can use the principle of conservation of angular momentum. According to this principle, the total angular momentum before and after an event remains constant, provided no external torques act on the system.

The initial angular momentum of the merry-go-round can be calculated using the formula:

L1 = I1 * ω1

where L1 is the initial angular momentum, I1 is the moment of inertia, and ω1 is the initial angular speed.

The angular momentum of the child can be calculated using the formula:

L2 = m * r * v

where L2 is the angular momentum of the child, m is the mass of the child, r is the radius of the merry-go-round, and v is the velocity of the child.

Since the child jumps onto the edge of the merry-go-round, the radius of the system remains constant. We can assume that the child jumps with no initial velocity, so v is zero.

Now, the total angular momentum after the child jumps onto the merry-go-round can be written as:

L_total = L1 + L2

Since the total angular momentum remains constant, we can equate the initial and final angular momenta:

L1 = L_total - L2

Rearranging the equation, we get:

L_total = L1 + L2

Now, we substitute the given values into the equations:

I1 = 411 kg m^2 (moment of inertia of the merry-go-round)
ω1 = 10.4 rev/min (initial angular speed of the merry-go-round)
m = 38.3 kg (mass of the child)
r = 1.92 m (radius of the merry-go-round)
v = 0 m/s (velocity of the child)

L1 = I1 * ω1
L2 = m * r * v

L_total = L1 + L2

Calculating L1:

L1 = 411 kg m^2 * (10.4 rev/min) = 4274.4 kg m^2/min

Calculating L2:

L2 = 38.3 kg * 1.92 m * 0 m/s = 0 kg m^2/min

Now, substituting the values into the equation:

L_total = 4274.4 kg m^2/min + 0 kg m^2/min = 4274.4 kg m^2/min

Since L_total = L1 + L2, the final angular momentum is also 4274.4 kg m^2/min.

To find the final angular speed (ω2), we can rearrange the equation L2 = I2 * ω2, where I2 is the moment of inertia of the system after the child jumps on.

I2 = I1 + m * r^2

Substituting the values:

I2 = 411 kg m^2 + (38.3 kg) * (1.92 m)^2
I2 = 411 kg m^2 + 143.6352 kg m^2
I2 ≈ 554.6352 kg m^2

Now, calculate the final angular speed (ω2):

L2 = I2 * ω2

ω2 = L2 / I2

ω2 = 4274.4 kg m^2/min / 554.6352 kg m^2 ≈ 7.7 rev/min

Therefore, the new angular speed of the merry-go-round after the child jumps on is approximately 7.7 rev/min.

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