Relative Speed Problem: What is Matt's speed from Mary's point of view?

AI Thread Summary
Mary is on a merry-go-round, and Matt observes her approaching at 1 m/s. The discussion centers on calculating Matt's speed from Mary's perspective, which is determined to be 2 m/s, contrary to initial assumptions that it should be 1 m/s. The confusion arises from the translation of the problem and the missing information about the radius of the merry-go-round, which affects the relative speeds. Ultimately, the correct interpretation involves considering the angular velocity of the merry-go-round, leading to the conclusion that the problem is valid and the answer of 2 m/s is accurate. The complexities of non-inertial reference frames are highlighted as a key factor in understanding the relative speeds.
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Homework Statement


Mary sits on the perimeter of a merry-go-round undergoing uniform circular motion. Matt stands on the ground 12 m from the centre of the merry-go-round. Matt observes Mary coming towards him at a speed of 1 m/s. What is Matt's speed from the point of view of Mary?

(Ans: 2m/s)

Homework Equations





The Attempt at a Solution


I made a sketch according to the given problem. A denotes Mary and B denotes Matt.

Shouldn't the answer be 1 m/s? If Matt observes Mary coming towards him with 1m/s, then Mary should also observe Matt coming towards her with 1 m/s. I don't see why this is wrong.

Any help is appreciated. Thanks!
 

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Neither do I.

Chet
 
Chestermiller said:
Neither do I.

Chet

I myself wanted to dismiss the answer as wrong but considering that the problem is from a very trusted source, I couldn't. Here is the source: http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=P4619&l=en

Is my interpretation of problem correct? Would it be right to say that 1m/s is along the tangent of merry go round?
 
Maybe it loses something in the translation. They are talking about speeds, not velocities. Who know's what they really meant?

Chet
 
Chestermiller said:
Maybe it loses something in the translation. They are talking about speeds, not velocities. Who know's what they really meant?

Chet

I think you are right. Usually the translation is ok but looks like they missed it this time.

I tried to look at the Magyar version of the website. This is the Magyar version of the problem: http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=P4619&l=hu . From Google Translate, I see that they mention the radius of merry-go-round which is absent from the English version. Maybe they missed something more?
 
I think they meant velocity by sebesség.

(Szerintem a sebesség velocity lenne angolul)

Also, I don't think that sugár is a brilliant replacement word for radius, but that's just my opinion.
 
TheAustrian said:
I think they meant velocity by sebesség.

(Szerintem a sebesség velocity lenne angolul)

Do you know Hungarian? If so, can you please translate the problem? That would be greatly appreciated. Thanks!
 
Pranav-Arora said:
Do you know Hungarian? If so, can you please translate the problem? That would be greatly appreciated. Thanks!

Yes I speak it fluently.

Mari egy 6 méter sugarú, egyenletesen forgó körhinta szélén ül. Máté a körhinta középpontjától 12 méterre a földön áll. Máté úgy látja, hogy Mari éppen feléje mozog 1 m/s sebességgel. Mekkora sebességgel mozgónak látja Mari Mátét?

Mary is on the edge of a uniformly rotating carousel with radius of 6 meters. Mate stands 12 meters away from the axis of the carousel on the ground. Mate sees as Mari approaches her with a velocity of 1 m/s. At what velocity does the motion of Máté appear to Mari?
 
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TheAustrian said:
Yes I speak it fluently.

Mari egy 6 méter sugarú, egyenletesen forgó körhinta szélén ül. Máté a körhinta középpontjától 12 méterre a földön áll. Máté úgy látja, hogy Mari éppen feléje mozog 1 m/s sebességgel. Mekkora sebességgel mozgónak látja Mari Mátét?

Mary is on the edge of a uniformly rotating carousel with radius of 6 meters. Mate stands 12 meters away from the axis of the carousel on the ground. Mate sees as Mari approaches her with a velocity of 1 m/s. At what velocity does the motion of Máté appear to Mari?

Thank you once again!

So looks like the missing piece of info was the radius of merry-go-round and the velocity part. But still, I don't see why the answer should differ from 1m/s.
 
  • #10
I suspect that there is a Chance that the English to Hungarian translation went wrong somewhere. (the Question on the Hungarian website is taken from an American paper)
 
  • #11
TheAustrian said:
I suspect that there is a Chance that the English to Hungarian translation went wrong somewhere. (the Question on the Hungarian website is taken from an American paper)

Ah, looks like I should accept that the problem is erroneous. :(
 
  • #12
There are many ways for a question to go wrong. Possibly the answer of 2 on the website is a typo. Who knows?
 
  • #13
TheAustrian said:
There are many ways for a question to go wrong. Possibly the answer of 2 on the website is a typo. Who knows?

The problem statement is probably erroneous because I doubt they would ask such an easy problem (with answer 1m/s) for 4 points. :-p
 
  • #14
There's a way to get an answer of 2 m/s. Note first that Mary sits on the edge of the merry-go-round.

What if she sat at the center of the merry-go-round? Would Mary see Matt moving at the same speed that Matt sees Mary moving?
 
  • #15
TSny said:
What if she sat at the center of the merry-go-round? Would Mary see Matt moving at the same speed that Matt sees Mary moving?

Why should it be different? :confused:
 
  • #16
If Mary sits at the center, then she sees Matt (and trees, houses, etc.) whirling around her.
 
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  • #17
TSny said:
If Mary sits at the center, then she sees Matt (and trees, houses, etc.) whirling around her.

Ah yes, I think I see how you reach the answer of 2m/s. If I consider the angular velocity to be 6 rad/s, then Matt has a speed of 2 m/s in frame of Mary. This is probably what the question meant. Thanks TSny! :smile:
 
  • #18
6 rad/s?
 
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  • #19
TSny said:
6 rad/s?

Woops, I meant 1/6 rad/s. :redface:
 
  • #20
OK. Sounds good.
 
  • #21
Yes what we have here is V=R/ω = R/(v/r)=(R/r)v = (12/6)*1 = 2 m/s. The problem is correct.

That's a tricky question.
 
  • #22
Those darn non-inertial ref. frames...
 
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