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 pres

Question

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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Thiên Thanh 6 months 2021-07-27T08:03:35+00:00 1 Answers 16 views 0

Answers ( )

    0
    2021-07-27T08:04:40+00:00

    Answer:

    Explanation:

    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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