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Time to traverse a circular arc?

  • Thread starter g.licata
  • Start date
  • #1
17
0

Homework Statement


A car is driven along straight section of track length (r). The car started from rest and the time it took to cover r distance is (t). If the same car is then tested on a circular arc of radius r, starting from rest and continues to speed up at the maximum possible rate that allows it to remain on the track, how long would it take to traverse the arc?


Homework Equations



V_avg = (V_f + V_0) / (2)
V_avg = r / t
a = (V_f - V_0) / t
a =V^2 / r

The Attempt at a Solution



V_0 = 0
V_avg = (V_f) / 2. V_f = 2 * V_avg
a = (2 * r) / t^2
a = (v^2) / r = (2 * r) / t^2
t^2 = (2 * r^2) / v^2
t = (r/v) * sqrt(2)

That is what I got for time, I think I'm doing it correctly, but my classmates have different answers for the time it took to traverse the arc. Is there an error in my work?

Thanks :)
 
Last edited:

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
I don't understand this! a = 2r/t² would be the acceleration on the straight section, assuming constant acceleration. a = v²/r is the centripetal acceleration when on an arc with radius r. Is r the radius or is it the length of the arc in this question? You have taken it to be both in writing
a = (v^2) / r = (2 * r) / t^2
and your "limit" is that the acceleration is the same as in the linear case.
The "maximum rate that allows it to stay on the track" would be dependent on the coefficient of friction, which is not given.
 
  • #3
17
0
Oh sorry, I somehow forgot to put in r is the radius of the circular arc.
 
  • #4
Delphi51
Homework Helper
3,407
10
Is r also the length of the arc?
Do you have the coefficient of friction? If not, you have no limit on the acceleration due to circular motion and you will get the same answer as on the straightaway.

Make sure you give the entire question - every word and picture - or it is very frustrating.
 

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