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 28 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 of grinding, what is the force with which the man presses the knife against the stone?

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jolis1796 jolis1796 · Answer 1 · 2020-02-08T09:27:07+01:00

Answer:

F = 2.57 N .

Explanation:

Given data:

Diameter of the stone: d = 35 c m = 0.35 m

Radius: r = 35/2 cm = 17.5 cm = 0.175 m

Initial angular velocity: ω₀ = 200 rpm = (200 )(2π/60) rad/s = 20.94 rad/s

Final angular velocity

ω f = (1.00-0.10)*(200 rpm) = 180 rpm = (180 )(2π/60) rad/s = 18.85 rad/s

Mass of the stone

m = 28 K g

Coefficient of kinetic friction: μ k = 0.2

Time: t = 10 s

Part a:

By using the equation of motion with uniform angular acceleration we will find out the angular acceleration of grindstone:

ω f = ω₀ + α*t

⇒ α = (ω f-ω₀)/t = (18.85 rad/s - 20.94 rad/s)/(10 s) = - 0.21 rad/s²

A negative sign shows that the wheel is decelerating.

Part b:

Consider the rim of grindstone as a solid disk, therefore, the moment of inertia

I = 0.5*m*r²

We know that:

Torque = Force * Perpendicular distance

τ = (μ k*F )* r (I)

Also:

τ = I *α = (0.5*28 Kg*(0.175 m)²)(0.21 rad/s²) = 0.09 N-m

Equation (I) becomes,

τ = (μ k*F )* r ⇒ F = τ / (μ k*r)

⇒ F = (0.09 N-m) / (0.20*0.175 m)* = 2.57 N

So, the tradesman pressing the knife against grindstone by 2.57 N .