To solve this problem, we can use the conservation of angular momentum. The total angular momentum before Ronald jumps onto the merry-go-round must equal the total angular momentum after he jumps onto it.
Before Ronald jumps, the angular momentum of the merry-go-round (L1) is 0 (since it is stationary), and the angular momentum of Ronald (L2) is given by L2 = m * r * v, where m is his mass, r is the distance from the axis of rotation, and v is his velocity. So,
L2 = (80 kg) * (1.50 m) * (4.00 m/s) = 480 kg*m^2/s
The total initial angular momentum (L_initial) is the sum of the two: L_initial = L1 + L2 = 0 + 480 kg*m^2/s = 480 kg*m^2/s
After Ronald jumps onto the merry-go-round, the angular momentum of the system is given by the moment of inertia of the merry-go-round (I) multiplied by its angular velocity (ω). Therefore,
L_final = I * ω
Now using conservation of angular momentum, we set the initial and final angular momenta equal:
L_initial = L_final
480 kg*m^2/s = (750 kg*m^2 + 80 kg * (1.50 m)^2) * ω
Now, we solve for ω:
ω = 480 kg*m^2/s / (750 kg*m^2 + 80 kg * (1.50 m)^2)
ω = 480 kg*m^2/s / (750 kg*m^2 + 80 kg * 2.25 m^2)
ω = 480 kg*m^2/s / (750 kg*m^2 + 180 kg*m^2)
ω = 480 kg*m^2/s / 930 kg*m^2
ω ≈ 0.516 m^2/s
Therefore, the angular velocity of the merry-go-round just after Ronald Johnson has jumped onto it is approximately 0.516 rad/s.
Ronald Johnson, who is 80.0 kg, is running at 4.00 m/s when he suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. Ronald Johnson is traveling tangentially to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 750 kg∙m2 and very little friction at its rotation axis. What is the angular velocity of the merry-go-round just after Ronald Johnson has jumped onto it?
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