= Solution
The second <moment> is
$$
\mathbb E[S^2]=\int_0^r\frac{s^2}{R}\,ds=\frac{r^3}{3R}.
$$
Subtracting the squared <expected value> gives the <variance> and <standard deviation>:
$$
\boxed{\operatorname{Var}(S)=\frac{r^3}{3R}-\frac{r^4}{4R^2}
=\frac{r^3(4R-3r)}{12R^2},\qquad
\sigma=\sqrt{\frac{r^3(4R-3r)}{12R^2}}.}
$$
The <variance> is nonnegative because $0<r\leq R$. Both $\mu$ and $\sigma$ have units of time. At $r=R$ the result reduces to $R/\sqrt{12}$, the <standard deviation> of a <uniform distribution> over one full cycle.
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