Answer:
[tex]n \approx 5.778\times 10^{-4}\,rev[/tex]
Explanation:
The Work-Energy Theorem is applied herein:
[tex]-\frac{1}{4}\cdot m_{cyl}\cdot R^{2}\cdot \omega^{2} = - \mu_{k}\cdot N\cdot r \cdot 2\pi \cdot n[/tex]
The number of turns needed to stop the grindstone is:
[tex]n = \frac{m_{cyl}\cdot R^{2}\cdot \omega^{2}}{4\cdot \mu_{k}\cdot N \cdot r\cdot 2\pi}[/tex]
[tex]n = \frac{(15\,kg)\cdot (0.186\,m)^{2}\cdot [(1.83\,\frac{rev}{min} )\cdot (\frac{2\pi\,rad}{1\,rev} )\cdot (\frac{1\,min}{60\,s} )]^{2}}{4\cdot (0.80)\cdot (8.82\,N)\cdot(0.186\,m)\cdot 2\pi}[/tex]
[tex]n \approx 5.778\times 10^{-4}\,rev[/tex]
