Required information
An object moving at constant speed $v$ around a circle of radius $r$ has an acceleration a directed toward the center of the circle. The SI unit of acceleration is $\mathrm{m} / \mathrm{s}^{2}$.
If the speed is increased $21.5 \%$, by what percentage does the radial acceleration increase?
$\%$
Answer
So the acceleration increases by $\boxed{47.2225}$ percent.
Step 1 ：Let the original speed be $v$ and the original radius be $r$. The original acceleration is $a = \frac{v^2}{r}$.
Step 2 ：The new speed is $\frac{121.5}{100}v$, so the new acceleration is $a' = \frac{(\frac{121.5}{100}v)^2}{r} = \left(\frac{121.5}{100}\right)^2 \frac{v^2}{r}$.
Step 3 ：This is $\left(\frac{121.5}{100}\right)^2$ times the original acceleration, which is an increase of $\left(\frac{121.5}{100}\right)^2 - 1 = \frac{14722.25}{10000} - 1 = \frac{4722.25}{10000}$.
Step 4 ：So the acceleration increases by $\boxed{47.2225}$ percent.
