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= 30kg are standing on opposite sides at a distance ro= 1.4m from the central axis. The merry-go-round is initially at rest. A person on the ground applies a constant tangential force of F= 2.5×102N at the rim of the merry-go-round for a time Δt= 12s . 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.

please tell the answers for this one --

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A playground merry-go-round has a radius of R= 4 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= 25kg are standing on opposite sides at a distance ro= 3.2m 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= 12s . For your calculations, assume the children to be point masses.

please tell the value for KE :(

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.