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

Staff Emeritus
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: Mar 14, 2005
3. Mar 14, 2005

harhar

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

4. Mar 14, 2005

HallsofIvy

Staff Emeritus
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

Staff Emeritus
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.