Question

A uniform cylindrical grindstone has a mass of 15.0 kg and a radius of 18.6 cm. The grindstone reaches a rotational velocity of 1.83 rev/min before its motor turns off. After the grindstone motor is turned off, a knife blade is pressed against the outer edge of the grindstone with a perpendicular force of 8.82 N. The coefficient of kinetic friction between the grindstone and the blade is 0.80. Use the work energy theorem to determine how many turns the grindstone makes before it stops.

Answer 1

Answer:

$n \approx 5.778\times 10^{-4}\,rev$

Explanation:

The Work-Energy Theorem is applied herein:

$-\frac{1}{4}\cdot m_{cyl}\cdot R^{2}\cdot \omega^{2} = - \mu_{k}\cdot N\cdot r \cdot 2\pi \cdot n$

The number of turns needed to stop the grindstone is:

$n = \frac{m_{cyl}\cdot R^{2}\cdot \omega^{2}}{4\cdot \mu_{k}\cdot N \cdot r\cdot 2\pi}$

$n = \frac{(15\,kg)\cdot (0.186\,m)^{2}\cdot [(1.83\,\frac{rev}{min} )\cdot (\frac{2\pi\,rad}{1\,rev} )\cdot (\frac{1\,min}{60\,s} )]^{2}}{4\cdot (0.80)\cdot (8.82\,N)\cdot(0.186\,m)\cdot 2\pi}$

$n \approx 5.778\times 10^{-4}\,rev$