Question

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 0.20. If the stone slows steadily to 180 rpm in 10 s of grinding, what is the force with which the man presses the knife against the stone?

Answer 1

Answer:

F = -2.205N

Explanation:

First, we have to find the angular aceleration due to the knife following the next equation:

W = Wo + at

where W is the final angular velocity and Wo is the initial angular velocity, a the angular aceleration and t the time.

Now, we will change the angular velocity to rad/s as:

Wo = 200 rpm = 20.94 rad/s

W = 180 rpm = 18.84 rad/s

Replacing in the previus equation, we get:

18.84rad/s = 20.94rad/s + a(10s)

solving for a:

a = -0.21rad/s^2

Now, we have to find the moment of inertia of the grindstone using:

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

Where M is the mass of the stone and R the radius of the stone. Replacing values:

I = $\frac{1}{2}(28kg)(0.15m)^2$

I = 0.315 kg*m^2

Adittionally:

T = Ia

where T is the torque, I the moment of inertia and a the angular aceleration.

so:

$U_kFd = Ia$

where $U_k$ is the coefficient of the kinetic friction, F is the force with which the man presses the knife and d the lever arm. So, replacing values, we get:

$(0.2)F(0.15m) = (0.315)(-0.21rad/s^2)$

solving for F:

F = -2.205N

it is negative because the stone is stopping due of this force.