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

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In summary, the conversation discusses the calculation of linear velocity and acceleration on a rotating merry-go-round. The equation for tangential velocity is derived and it is confirmed that tangential velocity is the same as linear velocity. The website provided helps clarify the concept of centripetal acceleration and angular acceleration. The conversation ends with a study tip for future courses.
  • #1
kaylanp01
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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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  • #2
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:
 
  • #3
You derived the equation, so just plug in your numbers.
 
  • #4
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.
 
  • #5
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 ;).
 
  • #6
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.
 
  • #7
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?
 
  • #9
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.
 

1. What is the purpose of a merry go round?

A merry go round is a type of amusement ride that rotates in a circular motion. Its purpose is primarily for entertainment and recreational purposes, especially for children. However, it can also be used for educational purposes to teach concepts such as linear speed and centrifugal force.

2. How is the linear speed of a merry go round calculated?

The linear speed of a merry go round can be calculated by dividing the distance traveled by the amount of time it takes to travel that distance. The distance traveled is the circumference of the merry go round, which can be calculated by multiplying the diameter by pi (π). The time can be measured using a stopwatch or by counting the number of revolutions in a certain period of time.

3. What factors affect the linear speed of a merry go round?

The linear speed of a merry go round can be affected by several factors, including the diameter of the ride, the rate at which it rotates, and the weight and distribution of the riders. The speed can also be affected by external factors such as friction and air resistance.

4. Is the linear speed the same for all points on a merry go round?

No, the linear speed is not the same for all points on a merry go round. The outer edge of the ride has a larger circumference and therefore travels a greater distance in the same amount of time compared to a point closer to the center. This means that the linear speed is faster at the outer edge and slower at the center.

5. How does the linear speed of a merry go round compare to other amusement rides?

The linear speed of a merry go round is generally slower compared to other amusement rides such as roller coasters or spinning rides. This is because the main purpose of a merry go round is not to provide thrill or speed, but rather a gentle and enjoyable experience. However, the linear speed can vary depending on the size and design of the ride.

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