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

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Homework Help Overview

The problem involves calculating the new angular speed of a merry-go-round after a child jumps onto it. The context includes parameters such as the radius, moment of inertia, and initial rotation speed of the merry-go-round.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of moment of inertia for the system, the conservation of angular momentum, and the conversion of angular speed between different units. Some question the accuracy of their calculations and the assumptions made regarding the system.

Discussion Status

There are multiple attempts to calculate the new angular speed, with participants providing different values and methods. Some guidance on using conservation of angular momentum is noted, but no consensus on the final answer has been reached.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may influence the level of detail and rigor in their calculations. There is an ongoing discussion about the correct interpretation of the moment of inertia and its impact on the final angular speed.

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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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Again, show your work.
 
How does this look?

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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