Question

need help ASAP please

A playground merry-go-round of radius R = 2.10 m has a moment of inertia of I = 260 kgm2 and is rotating at 11.0 rev/min about a frictionless vertical axis. Facing the axle, a 23.0 kg child hops on to the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round? O 0.229 rad/s 1.21 rad/s 0.829 rad/s 0.829 rad/s 2.95 rad/s

Answer 1

As we know that, know that , L= I xw. and initial angular speed (w;) is given by, 2nx (freg. of sot). = w;= 25x11 = 227 rad/min 1.152 rad/s. Then, initial angual momentum = 4; - IXw; = 260 x1152 Li = 299.4985 kg.me/s. Then, Extra moment of inertia added by kid (II) Ik = more = 23 x 2.12 = 101.43 kg.mt. Then, final moment of imestia (If) = 1; + Ik =2601101.43 = 361.43 kg.mt. Then, final angular yngmenta speed; we is given by, L 299.4985 0.8286487 rad/s. Is 361.43 Wf 60 Then, -7.913 rev/min wf = 0.8286487x 20

Final Answer: Option (c) or Option (d) 0.829 rad/s

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