How Do You Calculate Angular Motion for a Child on a Merry-Go-Round?

AI Thread Summary
To calculate angular motion for a child on a merry-go-round, the rotational inertia (I) of the system is determined using the formula I = 1/2mr^2, where m is the total mass. The angular velocity (ω) can be calculated using ω = 2θ/t, with θ representing the total angular displacement in radians. The child’s initial position affects the calculations, particularly when moving to the edge of the ride, which requires reevaluating the system's moment of inertia. The discussion highlights the importance of distinguishing between the moment of inertia for the disk and the child treated as a point mass. Understanding these principles is crucial for accurately solving the problem.
LizAnn10
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Homework Statement


Okay so this is a multi part question. I think I am on the right track, but I am most likely wrong. Any help is good. If I know what to use for the equations I can usually solve it on my own.
The question is:
A child of mass 40kg is on a merry go round. The ride is a uniform disk of mass 150kg and radius 2m. The child is hanging onto a pole at a distance of 1m. Her father, mass 80kg pushes on the ride with a constant tangential force of 100N.
A) Determind the rotational inertia of the system (child and ride)
B) Determind the angular velocity in rpm of the child after 5s
C) The man releases the ride and it continues at a constant angular velocity. If the child moves to the ouside of the ride, 2m away from the center, what will be the new angular velocity?


Homework Equations


A) This was the easy one. I=1/2mr^2
B) I think this one is θ=(ω/2)t (its really ωfinal plus ωinitial but the initial ω would be zero, so I canceled it out.
I doesn't specify the revolutions so I assumed that it was just one. (I hope that is right) But I think that would make θ=6.283rad
C) I have no clue. I feel like it would be the same as B but distance, the only thing that changes, isn't a factor. Also it doesn't give a new time or anything.

The Attempt at a Solution


A) 360kgm^2
B) 150rad/s if I did it right.
C) No clue
 
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LizAnn10 said:
A) This was the easy one. I=1/2mr^2

What about the child?
B) I think this one is θ=(ω/2)t (its really ωfinal plus ωinitial but the initial ω would be zero, so I canceled it out.
I doesn't specify the revolutions so I assumed that it was just one. (I hope that is right) But I think that would make θ=6.283rad

The question asks for ω, not θ.
C) I have no clue. I feel like it would be the same as B but distance, the only thing that changes, isn't a factor. Also it doesn't give a new time or anything.

Something has definitely changed; see part A.
 
Thanks for replying.
For A, I used the mass of the child plus the mass of the ride for m. Is that right?
For B, I just solved for ω, so it was really ω=2θ/t, so it was ω=2(6.283rad)/5s Is that right?
For C, I'm still not getting it..
 
LizAnn10 said:
For A, I used the mass of the child plus the mass of the ride for m. Is that right?
No. Disks and point masses (assuming the child as one) do not have the same moment of inertia.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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