Question

A 36.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates at 3.10 rad/s. If the child moves to the edge of the merry-go-round, what is the new angular velocity of the system? Model the merry-go-round as a solid disk. |

Answer #2

Mass of child = m = 36.5 kg

Mass of merry-go-round = M = 125 kg

Radius of the merry-go-round = R

Initial angular velocity of the merry-go-round =
_{1} = 3.1 rad/s

Final angular velocity of the merry-go-round =
_{2}

Moment of inertia of merry-go-round = I = MR^{2}/2

Speed of child at the edge of the merry-go-round = V =
_{2}R

By angular momentum conservation,

I_{1}
= I_{2}
+ mVR

(MR^{2}/2)_{1}
= (MR^{2}/2)_{2}
+ m(_{2}R)R

M_{1}/2
= M_{2}/2
+ m_{2}

(125)(3.1)/2 =
_{2}(125/2 + 36.5)

_{2}
= 1.957 rad/s

The new angular velocity of the system = 1.957 rad/s

answered by: anonymous

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