Angular Distance driven by a car going around a roundabout

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SUMMARY

The discussion centers on calculating angular displacement for a car navigating a roundabout, specifically addressing the equation (530/360) x 2π x 6.3 = 58.277 for angular distance. Participants clarify that angular displacement can be represented in multiple ways, such as 530 degrees or 530 - 360 = 170 degrees, emphasizing the distinction between angular displacement and linear displacement. The conversation highlights common misconceptions among students regarding the interpretation of angular measurements.

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Homework Statement
An inexperienced driver entered a roundabout, and could not figure out how to exit. The driver went around the roundabout 530 degrees before finding a way out. At any given time while in the roundabout, the driver was 6.30 m from the center of the roundabout. What was the distance covered while in the roundabout?
Relevant Equations
2πr
Can someone look at my solution to see if it is correct.
 

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Your calculation makes it look more complicated than it is, but yes, you got the right answer. Why did you not simply write a single equation:

(530/360)x2pix6.3 = 58.277
 
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phinds said:
Your calculation makes it look more complicated than it is, but yes, you got the right answer. Why did you not simply write a single equation:

(530/360)x2pix6.3 = 58.277
ah that makes sense. I always over complicate things. Thank you for your feedback.

Also, if I wanted to know the displacement of the inexperienced driver in the roundabout, would it simply be 530 - 2*360?
 
aatari said:
if I wanted to know the displacement of the inexperienced driver in the roundabout, would it simply be 530 - 2*360?
530-360 degrees would be the angular displacement about the centre of the roundabout. Why do you want to send it negative?
 
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haruspex said:
530-360 degrees would be the angular displacement about the centre of the roundabout. Why do you want to send it negative?
what he said (very small).jpg
 
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Thanks guys! very helpful.
 
haruspex said:
530-360 degrees would be the angular displacement about the centre of the roundabout. Why do you want to send it negative?
Why would someone think the displacement is equal to the given angle. It doesn't make any sense.
 
aatari said:
Why would someone think the displacement is equal to the given angle. It doesn't make any sense.
Are you suggesting that 530-360 (=170) degrees is NOT the angular displacement?
 
aatari said:
Why would someone think the displacement is equal to the given angle. It doesn't make any sense.
I'm not sure what your issue is.
We are dealing with angular displacement, not linear displacement. It is a bit different in that any given angle has an infinite number of representations. You could say the angular displacement is 530, or 530-360, or 530±360n; as angular displacements they are all the same, but as angular distances they are all different.
Generally one adopts a conventional range spanning 360 degrees, such as [0 , 360) or (-180 , 180], etc.
 
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phinds said:
Are you suggesting that 530-360 (=170) degrees is NOT the angular displacement?
well I think the displacement should be 530-360 = 170. But in a group chat some students are adamant that the displacement is simply the angle given - which in this case would be 530 degrees.
 
  • #11
haruspex said:
I'm not sure what your issue is.
We are dealing with angular displacement, not linear displacement. It is a bit different in that any given angle has an infinite number of representations. You could say the angular displacement is 530, or 530-360, or 530±360n; as angular displacements they are all the same, but as angular distances they are all different.
Generally one adopts a conventional range spanning 360 degrees, such as [0 , 360) or (-180 , 180], etc.
aha thank you for clarifying this. It makes sense.
 

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