Another Rotational Motion Question

[dismalice]

Homework Statement

A tradesman sharpens a knife by pushing it against the rim of a grindstone. The 35 cm diameter stone is spinning at 200 rpm and has a mass of 15 kg. The coefficient of kinetic friction between the knife and the stone is 0.20.

If the stone loses 10% of its speed in 10 s of grinding, what is the force with which the man presses the knife against the stone?

Homework Equations

SIgma F=ma
F=mu_k*N
A_c= (Mv^2/r)
I would assume that because the knife is pushed into the stone force should be counted as centripetal force.

The question asked for N for units and values

The Attempt at a Solution

200 rpm * 2pi/60s = 20.944 rad/s
V = 20.944*.175 = 3.6652 m/s
N= 9.8*15= 147N

Velocity With Speed Decay = 3.2987 m/s
F=147*.2= 29.4N

I can't think of anywhere to go from here, and any sort of help would be appreciated.

 

[Delphi51]

Try it top-to-bottom. The big picture is
Torque = moment of inertia x angular acceleration

If you fill in the details on each of those parts and play around with it a bit you should be able to solve for the applied force.