# What is a if the speed is doubled without changing the circle’s radius?

A particle rotates in a circle with centripetal acceleration a = 14 { m/s^{2}}.

Picture the particle at the instant it is at the top of the circle. one-half rotation later, it will have the same speed, but in the opposite direction. As acceleration is defined as the change in velocity per unit time, the acceleration is (twice the velocity) / (half the period).
To make that easier to write, I’ll shorten those to:
Ao = 2Vo / .5Po

Because I don’t like fractions in the denominator, I’ll use the following instead:
Ao = 4Vo / Po

If you double the radius with the same speed, your change in velocity stays the same, but your period is double. This leaves you with:
A = (4Vo) / 2*(Po)

Compare the two:
A / Ao = ((4Vo) / 2*(Po)) / (4Vo / Po)

A little algebra will quickly show that:
A = Ao / 2

Now, if you keep the radius the same but double the speed, your velocity change will double, AND your period will be cut in half:
A = 2*(4Vo) / .5*Po
A = 4*(4Vo) / Po

Compare A to Ao again:
A / Ao = (4*4Vo / Po)) / (4Vo / Po)

A little algebra will quickly show that:
A = 4 * Ao

To sum up:
If you double the radius at the same speed, the centripetal acceleration is halved.
If you double the speed at the same radius, the centripetal acceleration is quadrupled.