# Centripetal Acceleration

1. Mar 14, 2005

### harhar

For centripetal acceleration, what is the relationship between the radius and the period?

2. Mar 14, 2005

### HallsofIvy

Do you mean the relationship for fixed acceleration?

If an object is moving in a circle, at constant speed, we can write the vector equation as $\vec r= R cos(\omega t)\vec i+ R sin(\omega t)\vec j$ (ω is the angular speed- the object will do one full circle ($2\pi$ radians) in $\frac{2\pi}{\omega}$ seconds and so the period is: one full circle in $\frac{\omega}{2\pi}$ seconds: the period.

The velocity vector is $-R\omega sin(\omega t)\vec i+ R\omega cos(\omega t)\vec j$ and the acceleration vector is $-R\omega^2 cos(\omega t)\vec i- R\omega^2 sin(\omega t)$ which has length $\alpha = R\omega^2$.
That is $\omega= \sqrt{\frac{\alpha}{R}}$ and so the period is
$$T= \frac{\omega}{2\pi}= \frac{1}{2\pi}\sqrt{\frac{\alpha}{R}}$$

You can also solve that for R:
$$R= \frac{4\pi^2T^2}{\alpha}$$

Last edited by a moderator: Mar 14, 2005
3. Mar 14, 2005

### harhar

oh man...too advanced for me....

Can somone just tell me if as the radius increases the period is longer or something?

4. Mar 14, 2005

### HallsofIvy

Yes, the period increases as the square root of the radius: if the radius is 4 times as large, the period is twice as large.

5. Mar 14, 2005

### dextercioby

Yes,if the velocity (the linear/tangetial) is kept constant...

Daniel.

6. Mar 14, 2005

### HallsofIvy

It's a little spooky to see responses to responses while you are editing!

7. Mar 14, 2005

### dextercioby

Would u mind if i told u that i didn't look at your post...?Before correcting it...

Daniel.

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