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Merry-Go-Round physics homework

  • Thread starter bpw91284
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  • #1
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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:

Answers and Replies

  • #2
Gyroscope
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!?
 
  • #3
340
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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:
  • #4
67
0
Used conservation of angular momemtum and got the correct answer. Though, why is KE not conserved?

Thanks,
Brandon
 
  • #5
Gyroscope
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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