Question

A 2.8-m-diameter merry-go-round with rotational inertia 110 kg?m2 is spinning freely at 0.60 rev/s . Four 25-kg children sit suddenly on the edge of the merry-go-round.

a.) Find the new angular speed.

b.) Determine the total energy lost to friction between the children and the merry-go-round.

Answer #1

Part A:

When Children suddenly sits on it, Angular momentum will be conserved.

Li = Lf

Ii*wi = If*wf

Ii = initial moment of inertia = 110 kg-m^2

wi = 0.60 rev/sec

If = final moment of inertia = Ii + 4*Mc*r^2 = 110 + 4*25*1.4^2 = 306 kg-m^2

wf = ?

wf = wi*(Ii/If)

wf = 0.6*(110/306) = **0.216 rev/sec = 1.36 rad/sec (Check
the units and use whichever you need)**

**Part B**

Loss in KE will be

dKE = KEf - KEi

dKE = 0.5*If*wf^2 - 0.5*Ii*wi^2

wi = 0.6 rev/sec = 3.77 rad/sec

dKE = 0.5*306*1.36^2 - 0.5*110*3.77^2 = -498.72 J

(-ve sign means energy lost).

Please Upvote.

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