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.

Question

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.

(a) What is the angular acceleration α of the merry-go-round? (in rad/s2)

α=

(b) What is the angular velocity ωfinal of the merry-go-round when the person stopped applying the force? (in rad/s)

ωfinal=

(c) What average power Pavg does the person put out while pushing the merry-go-round? (in Watts)

Pavg=

(d) What is the rotational kinetic energy R.K.Efinal of the merry-go-round when the person stopped applying the force? (in kg⋅m2/s2)

R.K.Efinal=

Elena · Answer

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.

Anonymous · Answer

The power is wrong

Greco · Answer

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