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 .
