A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is spinning at 200 rpm and has a mass of 28 kg. The coefficient of kinetic friction between the knife and the stone is 0.2. The stone slows steadily to 180 rpm in 10 s of grinding.

a) What is the magnitude of the angular acceleration of the grindstone as it slows down?

b) With what force does the tradesman press the knife against the grindstone?

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

b) Assume the grind stone is a solid disk, its moment of inertia is

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

So the friction torque is

The friction force is

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

i think its c if i'm wrong

answer///isobars;