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29 November, 08:32

A disc-shaped grindstone with mass 50 kg and diameter 0.52m rotates on frictionless bearings at 850 rev/min. An ax is pushed against the rim (to sharpen it) with a normal force of 160 N. The grindstone subsequently comes to rest in 7.5 seconds. What is the co-efficient of friction between ax and stone?

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  1. 29 November, 10:38
    0
    0.48

    Explanation:

    m = mass of disc shaped grindstone = 50 kg

    d = diameter of the grindstone = 0.52 m

    r = radius of the grindstone = (0.5) d = (0.5) (0.52) = 0.26 m

    w₀ = initial angular speed = 850 rev/min = 850 (0.10472) rad/s = 88.97 rad/s

    t = time taken to stop = 7.5 sec

    w = final angular speed 0 rad/s

    Using the equation

    w = w₀ + α t

    0 = 88.97 + α (7.5)

    α = - 11.86 rad/s²

    F = normal force on the disc by the ax = 160 N

    μ = Coefficient of friction

    f = frictional force

    frictional force is given as

    f = μ F

    f = 160 μ

    Moment of inertia of grindstone is given as

    I = (0.5) m r² = (0.5) (50) (0.26) ² = 1.69 kgm²

    Torque equation is given as

    r f = I |α|

    (0.26) (160 μ) = (1.69) (11.86)

    μ = 0.48
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