# Uniform circular motion?

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1. Jul 6, 2016

### Hardikph

• Member advised to use the homework template for posts in the homework sections of PF.

How is Δθ in circle equals to angle in velocity vectors triangle?
I tried using simple geometry but I can't.

2. Jul 6, 2016

### Staff: Mentor

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?

3. Jul 6, 2016

4. Jul 6, 2016

5. Jul 6, 2016

### Delta²

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).

6. Jul 6, 2016

### Hardikph

Delta2 and Clara I appreciate your explaination but can you please describe some '90 degree usage'.

7. Jul 6, 2016

### Staff: Mentor

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°.

8. Jul 6, 2016

### Clara Chung

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.

9. Jul 8, 2016

### James R

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.