Explanation:
We are told to find the rotational speed needed for a 6.0 -nch grinding wheel to generate a linear speed of 4100fpm
To do so, we will use the given formula:
[tex]L=\frac{\pi dr}{12}[/tex]
So, we are to solve for r, given that
[tex]\begin{gathered} d=6.0 \\ L=4100fpm \end{gathered}[/tex]
So, if we make r the subject of the formula
[tex]r=\frac{12L}{\pi d}[/tex]
Thus
[tex]r=\frac{12\times4100}{\pi\times6}=2610.14rpm[/tex]
If we are to round up our answer to the nearest hundred, we will have the rotational speed to be 2600 rpm
