To find the answers to these questions, we'll need to use the principles of rotational motion and apply the relevant formulas.
(a) The angular acceleration α of the merry-go-round can be calculated using the formula:
α = τ / I,
where τ is the torque applied and I is the moment of inertia about the axis of rotation.
In this case, there is a torque applied at the rim of the merry-go-round, given by the force multiplied by the radius: τ = F * R.
Plugging in the values, we have:
τ = (2.5 × 10^2 N) * (2 m) = 5 × 10^2 N·m.
Now, we can calculate the angular acceleration α:
α = τ / I = (5 × 10^2 N·m) / (9 × 10^3 kg·m^2) = 5.6 × 10^-2 rad/s^2.
Therefore, the angular acceleration α of the merry-go-round is 5.6 × 10^-2 rad/s^2.
(b) The final angular velocity ωfinal can be obtained using the equation of rotational motion:
ωfinal = ωinitial + α * Δt,
where ωinitial is the initial angular velocity (which is zero in this case), α is the angular acceleration, and Δt is the time interval.
Plugging in the values, we have:
ωfinal = 0 + (5.6 × 10^-2 rad/s^2) * (12 s) = 0.67 rad/s.
Therefore, the angular velocity ωfinal of the merry-go-round when the person stopped applying the force is 0.67 rad/s.
(c) The average power Pavg can be calculated using the formula:
Pavg = τ * ωavg,
where τ is the torque applied and ωavg is the average angular velocity during the time interval.
We already know the torque τ applied at the rim: τ = (2.5 × 10^2 N) * (2 m) = 5 × 10^2 N·m.
To find the average angular velocity ωavg, we can use the formula:
ωavg = (ωfinal + ωinitial) / 2 = (0 + 0.67 rad/s) / 2 = 0.34 rad/s.
Now, we can calculate the average power Pavg:
Pavg = τ * ωavg = (5 × 10^2 N·m) * (0.34 rad/s) = 170 W.
Therefore, the average power Pavg that the person puts out while pushing the merry-go-round is 170 Watts.
(d) The rotational kinetic energy (R.K.Efinal) of the merry-go-round when the person stopped applying the force can be calculated using the formula:
R.K.Efinal = (1/2) * I * ωfinal^2,
where I is the moment of inertia about the axis of rotation and ωfinal is the final angular velocity.
Plugging in the values, we have:
R.K.Efinal = (1/2) * (9 × 10^3 kg·m^2) * (0.67 rad/s)^2 = 2 × 10^3 kg·m^2/s^2.
Therefore, the rotational kinetic energy R.K.Efinal of the merry-go-round when the person stopped applying the force is 2 × 10^3 kg·m^2/s^2.