Simple circular motion conceptual question

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Doubling the speed in circular motion requires the radius to increase to four times its original size to maintain the same acceleration. The period of the motion also changes, and the correct relationship is derived from the formula T = 2πR/V, leading to a new period of 4πR/v when speed is doubled. There was some confusion regarding the definitions of period and time, but the correct understanding is that period is measured in seconds. The discussion highlights the importance of accurately applying formulas in physics. Overall, the participant expresses satisfaction with their exam performance despite minor errors in understanding.
RJLiberator
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Homework Statement


[/B]
If the speed was doubled, what would the radius and period be to keep the same acceleration?

Homework Equations



a = v^2/R
T = 2piR/t

The Attempt at a Solution



So with my understanding it is as follows:
(2v)^2/R
This makes it 4v^2/x = a
since a=v^2/R

Setting this into a proportion
we get
4v^2*R=v^2*x
so 4R=x

Radius must be 4 times larger to support a doubling in speed.

T = 8piR/t

Correct?
 
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Looks like it's correct to me.
 
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Yahoooo! I did extremely well on this physics exam, many thanks to people like you on this forum.

Cheers.
 
RJLiberator said:
If the speed was doubled, what would the radius and period be to keep the same acceleration?
It depends what speed is referred to here. Is it the linear speed at the periphery or the angular speed? Maybe you could post the whole question, word for word.
 
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Unfortunately, I don't have the question as it was an exam question.

However, the "car" was going at constant speed on the outside (atan=0?) and the speed was doubled according to arad from my memory.
 
You also have a T = 2piR/t
and you think it changes to T = 8piR/t

But forgot to mention what it T is or what t is. Dimensionally, this looks suspicious.
 
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Hm.
T = period.
period = 2pir/time right?

When the r becomes 4R it becomes 8pi*R/t
the only part i am unsure about there is the denominator, t for time, if it is needed.
 
RJLiberator said:
Hm.
T = period.
period = 2pir/time right?

When the r becomes 4R it becomes 8pi*R/t
the only part i am unsure about there is the denominator, t for time, if it is needed.
The calculation of period will involve speed.
 
Yes, wouldn't the units be 2piR = meters and t = seconds
so m/s ?
 
  • #10
Period is measured in seconds.
 
  • #11
Ah... I see my error, I was under the impression that the definition of period was T = 2piR/t however, that is incorrect. The definition is actually
V = 2piR/T which becomes 2piR/V = T and in my case it would be 8piR/2v which simplifies to 4piR/v and the units becomes m/m/s which is just seconds.

Ah, that is unfortunate, I won't score 100% because i had the wrong definition in my head :p.

Thank you.
 
  • #12
So if the acceleration stays the same, and the speed while making a turn doubles, the radius has to become four times as big. And the time needed to complete the turn doubles.
 
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  • #13
Indeed. Thank you for that understanding. :)
 
  • #14
trying to help here but it seems most questions I know are quickly answered, from where I'm standing the only algebra was sqauring your two and moving your centripittal acceleration equation around, then you just plugged those values in for T, if you happen to confuse T equation and did everything else right I doubt your teacher will dock you too much
 
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  • #15
Thanks Jediknight. Yeah, I'm not too bothered, I'm very happy with how the exam went :D
 

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