Merry-Go-Round physics homework

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Homework Help Overview

The problem involves a merry-go-round with a specified radius and mass, which is initially rotated with constant angular acceleration. The discussion focuses on calculating the final rotational kinetic energy and the effect of additional mass on the angular velocity of the system.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of the moment of inertia for a uniform disk and the equations for rotational kinetic energy. There are attempts to apply conservation principles, particularly regarding angular momentum and kinetic energy.

Discussion Status

Some participants have provided guidance on the conservation of angular momentum and questioned the assumptions regarding kinetic energy conservation. There is an ongoing exploration of the correct application of formulas and the implications of forces exerted by the children on the merry-go-round.

Contextual Notes

Participants are discussing the implications of ignoring friction and the assumptions made about the system's behavior when additional mass is added. There are also references to potential errors in the application of equations related to kinetic energy.

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Homework Statement



A merry-go-round in a playground has a radius of about 2 m and a mass of about 400 kg. Starting from rest, a parent begins to rotate it with constant angular acceleration. After 5 s, it reaches its final angular velocity of about pi/2 rad/s, and continues to rotate at that angular velocity

a.
Treat the merry-go-round as a uniform disk (the moment of inertia of a uniform disk is mr2/2); the final rotational kinetic energy of the merry-go-round is about ___ J.

b.
5 children, each with a mass of about 20 kg, step on the merry-go-round and stay near its edge. As they step on it, its angular velocity drops to about ____ rad/s. Ignore friction.



Homework Equations



I_disk=0.5mr^2
KE_rotational=0.5*I*w^2


The Attempt at a Solution



For part a I did...

KE=(0.5)(0.5)(400)(2^2)(pi/2)^2=987 J

For part b I did...

KE_i=KE_f

987 = 0.5(100)(2^2)(w^2) + 0.5(0.5)(400)(2^2)(w^2)

and get w=1.3 but the answer is supposed to be 1.0. Is mr^2 the correct way to do the kid's intertia?
 
Last edited:
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Why do you think that kinetic energy will remain the same?

What physical quantity you are sure that will remain constant? Maybe angular momentum, who knows!?
 
Yes, think about what gyro is stating.
Also, your KE eqn is incorrect, there should be an omega squared. You seemed to take this into account in your soln. though.*edit* it is fixed now
 
Last edited:
Used conservation of angular momentum and got the correct answer. Though, why is KE not conserved?

Thanks,
Brandon
 
When the kids jump to the merry-go-round they exert forces on it. Forces cause change in kinetic energy, do not they?
 

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