A playground merry-go-round has a radius of R= 2 m and has a moment of inertia Icm= 9×103kg⋅m2 about a vertical axis passing through the center of mass. There is negligible friction about this axis. Two children each of mass m= 20kg are standing on opposite sides at a distance ro= 1.2m from the central axis. The merry-go-round is initially at rest. A person on the ground applies a constant tangential force of F= 2.0×102N at the rim of the merry-go-round for a time Δt= 18s . For your calculations, assume the children to be point masses.

R=2 m, Ic=9000 kg•m², r₀=1.2 m, m=20 kg, F=200 N, Δt=18 s.

I=Ic+2I₀=Ic+2m r₀² =
=9000+2•20•1.2²=9057.6 kg•m².

M=Iα
α=M/I=FR/I =200•2/9057.6 = 0.044 rad/s²

ω=α•Δt =0.044•18=0.792=0.8 rad/s

P=Fv=P ωR = 200•0.8•2 = 320 W.

KE=Iω²/2 = 9057.6•0.8²/2 = 2898.4 J.

The power is wrong

power is

KE=Iù²/2 = 9000•0.8²/2 = 2880 J.
(Whitout the children)

To solve these questions, we will use the principles of rotational dynamics.

(a) The torque τ acting on the merry-go-round can be calculated using the formula τ = Iα, where I is the moment of inertia and α is the angular acceleration. As there are two children standing on opposite sides, each with a mass of 20 kg, the total torque can be calculated as τ = 2mroα = 40 kg × 1.2 m × α.

The person on the ground applies a tangential force at the rim of the merry-go-round, which creates a torque. The formula for torque is τ = rF, where r is the radius and F is the force. So, we have τ = 2 m × 2 m × F = 4 mF.

Setting these two equations equal, we have 40 kg × 1.2 m × α = 4 mF.

Plugging in the given values, we have 40 kg × 1.2 m × α = 4 × (2.0 × 10^2 N).

Solving for α, we get α = (4 × 2.0 × 10^2 N) / (40 kg × 1.2 m).

Calculating this, we find α = 3.33 rad/s^2.

(b) The final angular velocity ωfinal can be calculated using the formula ωfinal = ωinitial + αΔt, where ωinitial is the initial angular velocity and Δt is the time. Since the merry-go-round is initially at rest, ωinitial is 0 rad/s. Plugging in the given values, we have ωfinal = 0 rad/s + (3.33 rad/s^2) × (18 s). Calculating this, we find ωfinal = 59.94 rad/s.

(c) The average power Pavg can be calculated using the formula Pavg = τω/Δt, where τ is the torque, ω is the angular velocity, and Δt is the time. Plugging in the given values, we have Pavg = (4 mF) × (59.94 rad/s) / (18 s). Calculating this, we find Pavg = 79.92 W.

(d) The rotational kinetic energy R.K.Efinal can be calculated using the formula R.K.Efinal = (1/2) I ω^2, where I is the moment of inertia and ω is the angular velocity. Plugging in the given values, we have R.K.Efinal = (1/2) × (9 × 10^3 kg·m^2) × (59.94 rad/s)^2. Calculating this, we find R.K.Efinal = 3.59 × 10^6 kg·m^2/s^2.