Question

A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50.0 kg is rotating at 800 rev/min. You press an axe against the rim with a normal force of 160 N (see the figure below), and the grindstone comes to rest in 7.00 S. (Figure 1) Part A Find the coefficient of kinetic friction between the axe and the grindstone. There is negligible friction in the bearings Express your answer to three significant figures. Templates Symbols uñdo redp fadet keyboard shortcuts help Submit Request Answer < Return to Assignment Provide Feedback Figure 1 of 1 m = 50.0 kg 16ON

Answer 1

For a solid disk, radius of gyration, k = R/ √2

In this case, R = 0.52 /2 = 0.26

k = 0.26/√2 = 0.1838m

Inertia of disk = mk²

= 50 x 0.1838^2 = 1.689 kg.m²

Initial angular velocity, ω_i = 800 x 2 x π /60 = 83.78 rad/s

Deceleration : (ω_i - ω_f) /t = (83.78 - 0) / 7 = 11.97 rad/s²

Braking Torque = I α = 1.689 x 11.97 = 20.22 Nm

Torque = force x radius

20.22 = Friction force x 0.26

Friction force = 20.22 / 0.26 = 77.77 N

Coefficient of friction = F/ R = 77.77 /160 = 0.486

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