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9 November, 14:56

You have a horizontal grindstone (a disk) that is 85 kg, has a 0.31 m radius, is turning at 88 rpm (in the positive direction), and you press a steel axe against the edge with a force of 21 N in the radial direction.

a) assuming the kinetic coefficient of friction between steel and stone is. 2, calculate the angular acceleration of the grindstone in rad/s^2

b) how many turns will the stone make before coming to rest

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  1. 9 November, 17:38
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    moment of inertia of wheel

    I = 1/2 mr², m is mass and r is radius of disk.

    =.5 x 85 x. 31²

    = 4.08425 kg m²

    88rpm

    = 1.467 x 2π rad / s

    = 9.21 rad / s

    force of friction acting = 21 x. 2

    = 4.2 N

    torque produced by friction to stop it

    = friction x radius

    = 4.2 x. 31

    = 1.302 Nm

    angular deceleration = Torque / moment of inertia

    = 1.302 / 4.08425

    =.31878 rad / s²

    angular deceleration = -.31878 rad / s²

    b)

    initial angular speed ω₁ = 9.21

    Final angular speed ω₂ = 0

    angle of turn = θ radian

    ω₂² = ω₁ ² - 2 αθ, α is angular deceleration

    0 = 9.21² - 2 x. 31878 θ

    θ = 133.04 radian

    no of turns = 133.04 / 2π

    = 21.18 turns.
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