What is the relationship between radius and period for centripetal acceleration?

[harhar]

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

 

[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}

 

[harhar]

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

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

 

[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.

 

[dextercioby]

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

Daniel.

 

[HallsofIvy]

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

 

[dextercioby]

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


Daniel.