# Answer out by 2 orders of magnitude...how?

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1. Jun 16, 2015

### kostoglotov

1. The problem statement, all variables and given/known data

A tradesman sharpens a knife by pushing it against a grindstone with a given applied force. The grindstone is 28kg, 15 cm in radius, is originally going at 200 rpm but is slowed down to 180 rpm over 10 s by applying the knife. The coefficient of kinetic friction is 0.20. What is the applied force.

2. Relevant equations

$$\alpha = \frac{\Delta w}{\Delta t}$$

$$\tau = \alpha I$$

$$F_{net} = \frac{\tau}{r} = 0.20 F_{app}$$

$$I = \frac{1}{2}MR^2$$

$$\frac{rad}{s} = rpm\frac{2\pi}{60}$$

3. The attempt at a solution

$$\alpha = \frac{-20}{10} = -2 \frac{rpm}{s}$$

$$-2 \frac{rpm}{s} = \frac{-\pi}{15}\frac{rad}{s^2}$$

$$I = \frac{1}{2}28(0.15)^2 = 3.15\times 10^{-3} kg.m^2$$

$$\tau = \left|\frac{-\pi}{15}\right|3.15\times 10^{-3} \approx 6.6\times 10^{-3} N.m$$

$$F_{net} = \frac{6.6\times 10^{-3}}{0.15} = 4.4\times 10^{-3} N$$

$$F_{app} = \frac{4.4\times 10^{-3}}{0.20} = 0.022 N$$

2. Jun 16, 2015

### Nathanael

To make it easier on yourself (and others like me) you should not plug in values until the very end of the calculation. This makes it MUCH easier to trace back any mistakes. Also, the form of the final equation can sometimes give you insight.

Anyway your calculation for the moment of inertia is small by a factor of 100 for some reason. 14*0.15^2 is 0.315 not 0.00315