Relationship between velocity, acceleration, and a circle?

AI Thread Summary
The discussion focuses on the relationship between velocity, acceleration, and circular motion. It begins by establishing the perimeter of a circle and the speed of an object moving around it. A calculation error is identified in determining the change in velocity (Δv), which is corrected to reflect positive values. The conversation highlights the importance of constant acceleration assumptions in motion equations, revealing discrepancies in the calculations. The thread concludes by emphasizing the interconnectedness of position, velocity, and acceleration in understanding circular motion dynamics.
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The perimeter of a circle is 2πR (R=radius). [ref]
Acceleration = Δv/Δt (v=velocity, t=time). [ref]
Motion mathematics can always be reduced to multiple independent one-dimensional motions. [ref]
The distance an object travels while accelerating = vit + at2/2 (a=acceleration, vi=initial velocity). [ref]
  1. If a circle centered at (0,0) has a radius of 2m, then it has a diameter of 4m and a perimeter of 4π m.
  2. If an object is moving clockwise around this circle once every 4 seconds, then that object has a speed of 1π m/s.
  3. At the top of the circle, the object has an (x,y) velocity of (π,0); let this be t0.
  4. At the right of the circle, the object has an (x,y) velocity of (0,-π); let this be t1.
  5. Δt = t1-t0 = 1 second.
  6. Δv = v1-v0 = (π,0) - (0,-π) = (-π,-π).
  7. a = (-π,-π) / second (for that particular time interval).
  8. So displacement on the X-axis = π m/s * 1 s + ( (-π m/s2) * (1s)2 / 2 ) = π m + -π/2 m = π/2 ...
But the object traveling the perimeter of the 2m-radius circle every 4 seconds should be at (2,0) 1 second after (0,2). What am I doing wrong - displacement, acceleration, or something else/more?

Thanks.

(other references)
- velocity calculator
- kinematic equations
 
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Δv = v1-v0 = (π,0) - (0,-π) = (-π,-π).
This is wrong.
Δv= (π,π). By basic arithmetic, your pies are positive. 0 - - π = π, and π - 0 = π
 
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stvrbbns said:
8. So displacement on the X-axis = π m/s * 1 s + ( (-π m/s2) * (1s)2 / 2 ) = π m + -π/2 m = π/2
You can only use the equation ## s = ut + \frac 12 a t^2 ## when acceleration is constant.
 
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Karmaslap said:
Δv = v1-v0 = (π,0) - (0,-π) = (-π,-π).
This is wrong.
Δv= (π,π). By basic arithmetic, your pies are positive. 0 - - π = π, and π - 0 = π
Thanks for catching that. I think that it should then be:
Δv = v1-v0 = (0,-π) - (π,0) = (-π,-π)
and that I had the v0 and v1 incorrectly switched; the outcome is correctly still (-π,-π).
 
MrAnchovy said:
You can only use the equation ## s = ut + \frac 12 a t^2 ## when acceleration is constant.
Ah, I was trying to take the average acceleration and treat it as constant.
 
Plots of the actual component numbers show just how off an assumption of constant acceleration for each component is for this problem.

PsIq8v2.png
 
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spamanon said:
Plots of the actual component numbers show just how off an assumption of constant acceleration for each component is for this problem.
Thank you very much! That relationship between position, velocity, and acceleration exposes quite a bit of my problem.

[ref] For anyone reading this post later, acceleration is the derivative of velocity and velocity is the derivative of position. The graphs posted by @spamanon show all 3 of those for my circle (red position, blue velocity, black acceleration).
 
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