Calculating Angular Speed of a Merry-Go-Round After Adding a Child

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To calculate the new angular speed of a merry-go-round after a child jumps on, the moment of inertia and initial angular speed must be considered. The initial moment of inertia is 250 kgm², with the merry-go-round rotating at 15.0 rev/min, which converts to approximately 1.57 rad/s. After the child with a mass of 35 kg hops on, the new moment of inertia becomes 390 kgm². Using the conservation of angular momentum, the final angular speed is calculated to be about 9.62 rev/min. This demonstrates how adding mass affects the rotational speed of the system.
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Homework Statement



A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kgm2 and is rotating at 15.0 rev/min about a frictionless vertical axle. Facing the axle, a 35.0 kg child hopes onto the merry-go-round. What is the new angular speed of the merry-go-round?
 
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Ok so moment of intertia =35kg(2^2)=140
15rev/m*pi/30=1.57 rad/s

Linitial=Lfinal
Linitial=I*angular speed
250(1.57)=263.12w
392.5=263.12w
1.49 rad/s=w

Converting back to rev/min:
(1.49*60)/2pi=14.23 rev/min
 
bump.
 
I think this now is the correct answer.
Final moment of inertial (if)= 250kgm^2+(35kg)(2m)^2=390kgm^2

IiWi=IfWf
(250)(15)=390wf
3750=390wf
= 9.62 rev/min
 

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