A tradesman sharpens a knife by pushing it against the rim of a grindstone. The

Question

A tradesman sharpens a knife by pushing it against the rim of a grindstone. The 45 cm diameter stone is spinning at 200 rpm and has a mass of 24 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?

A tradesman sharpens a knife by pushing it against the rim of a grindstone. The 45 cm diameter stone is spinning at 200 rpm and has a mass of 24 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?

Explanation / Answer

Change in angular momentum = i w

    = 10% of (0.5MR^2 w)

    = 0.1*(0.5*24*(0.45/2)^2 *(200*2pi/60)

     = 1.2723 kgm^2/s

Torque = angular momentum change/time

     = 1.2723/10 = 0.1272 Nm

Torque = F*R

Tor is causwd by friction,

so Friction Force F = 0.1272/(0.45/2)

     = 0.565 N

Normal = Friction /u = 0.565/0.20

=2.825 N answer

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