91Ó°ÊÓ

the determinant of the drill bit $D=\left(12.7mm\right)\left(\frac{1}{1000mm}\right)$.

$=12.7×{10}^{-3}m$

The radius of the drill bit is,

role="math" localid="1667977368017" $$\begin{gathered} r=\frac{D}{2}=\frac{12.7}{2}×{10}^{-3}m \\ =6.35×{10}^{-3}m \end{gathered}$$

Now, the Angular speed of the drill bit,

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

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.

$v_{max}=r\mathrm{Ó\neg }$

Put the value here.

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

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

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,

role="math" localid="1667976803031" ${}^{a_r=\frac{v^2}{r}={\mathrm{Ó\neg }}^{2}r}$

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

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

$a_{r,max}={\mathrm{Ó\neg }}^{2}r$

Put the value $\mathrm{Ó\neg }$ and r,

We get,

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

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