Table 2.9.5(a) lists the information given in the statement of the example. The unknown distance from the center of the disk is $r$.

From the inequality $∥{\mathbf{F}}_{\mathbf{N}}∥$ = $m \mathrm{\κ} v^2\le d\/9$ lbs, obtain $\left(d\/g\right) \mathrm{\κ} v^2\le d\/9$, or, by making the appropriate substitutions from Table 2.9.5(a),

$\left(d\/32\right) \left(1\/r\right) {\left(3 \mathrm{\π} r\/2\right)}^2\le d\/9$ 

Consequently, $r\le \frac{128}{81 {\mathrm{\π}}^2}\=0.1601124877$ ft, or $\left(\frac{128}{81 {\mathrm{\pi }}^2}\cdot 12\right)\=1.921349852$ inches.