What is the Linear Speed of a Child on a Rotating Merry-Go-Round?

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Homework Help Overview

The problem involves calculating the linear speed of a child on a rotating merry-go-round, which completes one revolution in 3.80 seconds. The child is seated 1.24 meters from the center, and the discussion revolves around understanding the relationship between angular and linear velocity in the context of circular motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between distance from the center and velocity, with some questioning how to apply the derived equations for linear velocity. Others explore the concept of tangential velocity and its equivalence to linear velocity.

Discussion Status

The discussion is ongoing, with participants providing guidance on understanding the equations and concepts involved. Some express confusion about acceleration and its relation to circular motion, while others clarify the definitions and implications of centripetal acceleration.

Contextual Notes

Participants mention the need to pass course components and express concerns about their understanding of the material, indicating a focus on exam preparation and the challenges faced in grasping the concepts of circular motion and acceleration.

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Homework Statement


A rotating merry-go-round makes one complete revolution in 3.80 s. What is the linear speed of a child seated 1.24 m from the center?


Homework Equations


vt = wr
w = 2(pi)/T
Vt = 2(pi)r/T


The Attempt at a Solution


I know that the further you are from the center, the higher velocity that you'll have, but I'm not sure where to go with that.
 
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kaylanp01 said:
A rotating merry-go-round makes one complete revolution in 3.80 s. What is the linear speed of a child seated 1.24 m from the center?

Hi kaylanp01! :smile:

hmm :rolleyes: … let's try an easy case …

how far does the child go in 3.80s? :wink:
 
You derived the equation, so just plug in your numbers.
 
LowlyPion said:
You derived the equation, so just plug in your numbers.

So that's the equation for linear velocity? I'm confused. I just grabbed that out of my notes.
 
tiny-tim said:
Hi kaylanp01! :smile:

hmm :rolleyes: … let's try an easy case …

how far does the child go in 3.80s? :wink:

Wellll...technically...he doesn't go anywhere, as he ends up in the same place ;).
 
kaylanp01 said:
So that's the equation for linear velocity? I'm confused. I just grabbed that out of my notes.

Well your notes were right.

The trick now is to understand what you took notes about then isn't it.
 
LowlyPion said:
Well your notes were right.

The trick now is to understand what you took notes about then isn't it.
I'm just trying to get through it now...try to pass the CAPA part of the course even if I can't pass the exam. haha. It says in my notes that the equation I derived above was for tangential v...so are tangential and linear the same thing?
 
kaylanp01 said:
I'm just trying to get through it now...try to pass the CAPA part of the course even if I can't pass the exam. haha. It says in my notes that the equation I derived above was for tangential v...so are tangential and linear the same thing?

Yes tangential V is linear V.
 
  • #10
LowlyPion said:
Yes tangential V is linear V.

Okay, great.
I looked at that website, and it eliminates my next question (I was going to ask how to calculate acceleration, but its all there) so you just saved me (and yourself) from more pain. haha.
Thank you (twice). :)
 
  • #11
LowlyPion said:
Yes tangential V is linear V.
Okay. I've been defeated. I still can't get acceleration.
I used a = 2.05/1.24 = 1.65 and that's not correct.
 
  • #12
The acceleration is the centripetal acceleration v2/r
 
  • #13
LowlyPion said:
The acceleration is the centripetal acceleration v2/r

So how will I know when its centripetal and when it isn't? I'm so confused. :(
 
  • #14
kaylanp01 said:
So how will I know when its centripetal and when it isn't? I'm so confused. :(

If it is undergoing circular motion its mass is undergoing centripetal acceleration.

When the disk is changing ω at some rate of increase that is called α, and it is called angular acceleration.
 
  • #15
LowlyPion said:
If it is undergoing circular motion its mass is undergoing centripetal acceleration.

When the disk is changing ω at some rate of increase that is called α, and it is called angular acceleration.


Okay, its a little more clear now. I'm going to go study until my eyes water and/or begin to blur. Any study tips to offer?
 
  • #16
kaylanp01 said:
Okay, its a little more clear now. I'm going to go study until my eyes water and/or begin to blur. Any study tips to offer?

Next term read ahead before the lectures.
 
  • #17
LowlyPion said:
Next term read ahead before the lectures.
Thankfully, I won't be doing another physics course after this term. I usually read ahead throughout the semester, but really got off track within the past few weeks. Luckily, I currently have an A, so I can afford to do not-so-well on the final. ha.
 
  • #18
Good luck on the exam then.

Cheers.
 

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