Is Δθ in a circle equal to the angle in a velocity vectors triangle?

AI Thread Summary
The discussion centers on the relationship between the angle Δθ in a circle and the angle in a velocity vectors triangle. It explores the geometric implications of moving a radius vector r' slightly from r and how this affects the angle between the corresponding velocity vectors. The participants emphasize that the velocity vectors are always 90° rotated compared to the radius vectors, establishing a direct correlation between their angles. A mathematical relationship is derived showing that Δθ in the circle equals the angle in the velocity vectors triangle. The conversation highlights the importance of accurate vector representation and the geometric relationships involved.
Hardikph
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How is Δθ in circle equals to angle in velocity vectors triangle?
I tried using simple geometry but I can't.
 
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Try this:

What if r' were right on top of r? What is the angle between them? What is the angle between the velocity vectors?

Now move r' just a bit, an angle Δθ from r. How does the angle between the velocity vectors change?
 
Its because they are angles that have sides vertical to each other. r' is vertical to v' and r is vertical to v so the angle of (r',r) equals the angle of (v',v).
 
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Delta2 and Clara I appreciate your explanation but can you please describe some '90 degree usage'.
 
Another way to look at it: Realize that the velocity vectors are always 90° rotated compared to the r vectors. So if the angle that an r vector makes with the x-axis is θ, then the corresponding velocity vector must make an angle of θ + 90°.
 
Hardikph said:
Delta2 and Clara I appreciate your explanation but can you please describe some '90 degree usage'.

In the photo, θ+a = 90..(1) , a+b=90...(2), (1)-(2), θ=b , so it proves your statement "Δθ in circle equals to angle in velocity vectors triangle".
I hope it helps.
 
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Hardikph said:

One other point: the vector diagram on the right does not correspond to the vectors shown on the left.
Recall that ##\Delta \vec{v} = \vec{v}'-\vec{v}##. You also seem to have swapped the ##\vec{v}'## and the ##\vec{v}## in the diagram.
 
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