A playground merry-go-round has radius 2.60 m…?

Question

A playground merry-go-round has radius 2.60 m and moment of inertia 2300 kg*m^2 about a vertical axle through its center, and it turns with negligible friction.(a) A child applies an 23.0 N force tangentially to the edge of the merry-go-round for 21.0 . If the merry-go-round is initially at rest, what is its angular speed after this 21.0 s interval? w=?(b) How much work did the child do on the merry-go-round?W=?(c) What is the average power supplied by the child?P=? PHYSICS

Hermina Ferry · Accepted Answer

Torque τ = I α where α is angular accelerationTorque τ = F*r F*r = I α = I (ωf – ωi)/t =============================A) Given ωi = 0 F*r = I ωf/t ωf = F*rt/ I = 23*2.6*21/2300 = 0.546 rad/s ————————————–…B)W =0.5*I* ωf² = 1150*0.546 ² = 342.8 J —————–C)Power Total work/time =342.8334/21 =16.3W=================

Hermina Ferry · Answer

torque applied= 23 * 2.6= 59.8 Nm= moment of inertia * angular acc= 2300 *aa = 0.026 rad/s^2t =21; u=0v = u +atv = 0 + 0.026 * 21= 0.546 rad /sanswerangular speed after this 21.0 s intervalwork the child did on the merry-go-round= rotational KE gained = 0.5 * MoI * w^2= 0.5 * 2300 * 0.546^2= 342.83 Janswer average power supplied by the child= work the child did on the merry-go-round / time= 342.83 / 21= 16.33 Wanswer

Jakob Littel

A playground merry-go-round has radius 2.60 m…?

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