91Ó°ÊÓ

the determinant of the drill bit.

$$\begin{gathered} D=(12.7mm)\left(\frac{1}{1000mm}\right) \\ =12.7×{10}^{-3}m \end{gathered}$$

The radius of the drill bit is,

$$\begin{gathered} r=\frac{D}{2}=\frac{12.7}{2}×{10}^{-3}m \\ =6.35×{10}^{-3}m \end{gathered}$$

Now, the Angular speed of the drill bit,

role="math" localid="1664350970352" $$\begin{gathered} \mathrm{Ó\neg }=\left(1250\frac{rev}{min}\left(\right)\left(\frac{2\mathrm{Ï€}rad}{1rev}\right)\left(\frac{1min}{60s}\left(\right)\right)\right) \\ =\frac{125}{3}\mathrm{Ï€}rads \end{gathered}$$

angular speed is related to translation speed through relation,

$v=r\mathrm{Ó\neg }$

bits found in the circle r from the center of the bit have a high line speed.

$v_{max}$

Put the value here.

$v(6.35×{10}^{-3}m){\left(\frac{125}{3}\mathrm{π}\frac{rad}{s}\right)}_{max}$

$=0.831\frac{m}{s}.$

the maximum linear speed is$0.831\frac{m}{s}$.

a large part of the acceleration line$\stackrel{→}{a}$point in a solid rotating body. At a direct distance r from the rotating axis,

$a_r=\frac{v^2}{r}={\mathrm{Ó\neg }}^{2}r$

Here $\mathrm{Ó\neg }$is the magnitude.

the parts of the bit located on the circumference a distancefrom the center of the bit have the maximum radical speed.

$a{2}_{r,max}$

Put the value$\mathrm{Ó\neg }$and,

We get,

$a{\left(\frac{125}{3}\frac{\mathrm{π}rad}{s}\right)}^2{(6.35×{10}^{-3}m)}_{r,max}=109m/s^2$

The maximum radical acceleration is$109m/s^2$