Finding the Angular Velocity of a Merry-Go-Round

[PeachBanana]

Homework Statement

A person of mass 71 kg stands at the center of a rotating merry-go-round platform of radius 3.2 m and moment of inertia 920 kg * m^2 . The platform rotates without friction with angular /velocity 1.7 rad./s. The person walks radially to the edge of the platform.

Homework Equations

ω^2 * r = α

The Attempt at a Solution

The first question I asked myself was, "How long did it take him to walk to the edge of the platform?"

I found α to be ≈ 29.59 rad./s but I'm having trouble finding an equation relating this to time. I don't know θ and don't know ω final. Is there another way I should be looking at this?

 

[cepheid]

Homework Statement

A person of mass 71 kg stands at the center of a rotating merry-go-round platform of radius 3.2 m and moment of inertia 920 kg * m^2 . The platform rotates without friction with angular /velocity 1.7 rad./s. The person walks radially to the edge of the platform.

You haven't stated here what the problem is.

 

[PeachBanana]

I guess that would be a slight problem! The question being asked:

Calculate the angular velocity when the person reaches the edge.

 

[cepheid]

I think if you just compute the change in moment of inertia due to having the person on the edge rather than at the centre of rotation, then you can use conservation of angular momentum to find the answer.

 

[PeachBanana]

Okay, that makes sense. I think I calculated "I" incorrectly.

I said the initial moment of inertia was 920 kg * m^2. Then I thought the final "I" value would be (I assumed this merry-go-round was a solid cylinder) 1/2 (3.2 m)^2 (71 kg) ≈ 363.5 kg * m^2.

L initial = (920 kg * m^2)(1.7 rad./s)
L initial ≈ 1564 kg * m^2/s

1564 kg * m^2/s = 363.52 m^2 * kg * ω
ω ≈ 4.30 rad./s

 

[cepheid]

Okay, that makes sense. I think I calculated "I" incorrectly.

I said the initial moment of inertia was 920 kg * m^2. Then I thought the final "I" value would be (I assumed this merry-go-round was a solid cylinder) 1/2 (3.2 m)^2 (71 kg) ≈ 363.5 kg * m^2.

I'm quite puzzled by what you are attempting here. The moment of inertia of the merry-go-round itself is given in the problem. It's 920 kg m². You do not have to compute it. So why are you trying to?

Also, if the calculation you posted was supposed to be a calculation of the moment of inertia of the merry-go-round, then why did you use the mass of the person in the calculations?

What you have to do is find the change in the moment of inertia of the overall system (merry-go-round + person) given that the person moves from the centre of rotation to the edge. For this purpose, I think you can probably treat the person as a point mass.

 

[PeachBanana]

Okay. I understand that much better. I was trying to calculate the change in the moment of inertia but went about it the completely wrong way.