Question

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?

Answers

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.

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