A grinding wheel is in the form of a uniform solid disk of radius 6.99 cm and mass 2.08 kg. It starts from rest and accelerates uniformly un

Question

A grinding wheel is in the form of a uniform solid disk of radius 6.99 cm and mass 2.08 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.591 N · m that the motor exerts on the wheel.(a) How long does the wheel take to reach its final operating speed of 1 230 rev/min? s(b) Through how many revolutions does it turn while accelerating? revNeed Help?

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Latifah 3 years 2021-07-27T09:07:57+00:00 1 Answers 225 views 0

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  1. haidang
    0
    2021-07-27T09:09:54+00:00
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    Answer:

    A) 1.11 secs

    B) 11.41 revolutions

    Explanation:

    A) We are given final operating speed; ω_f = 1230rev/min = 1230 x 2π/60 rad/s = 128.81 rad/s

    Torque; τ = 0.591 N

    Mass;m = 2.08 kg

    radius; r = 6.99 cm = 0.0699m

    Let’s now find the acceleration from;

    torque = inertia x acceleration

    Inertia = (1/2)mr²

    Thus;

    τ = (½)(mr²)a

    So,

    0.591 = ½(2.08 x 0.0699²)a

    0.591 = 0.00508a

    a = 116.34rad/s²

    using the equation of rotational motion;

    ω_f = ω_i + at

    ω_i is initial angular velocity and it’s zero.

    Thus;

    128.81 = 0 + 116.34t

    128.81 = 116.34t

    t = 1.11secs

    B) the number of revolutions will be calculated from;

    θ = (ω_i)t + ½at²

    θ = 0(1.81) + ½(116.34)(1.11)²

    θ = ½(116.34)(1.11)²

    θ = 71.671257 rad

    Let’s convert it to revs

    θ = 71.671257 ÷ 2π

    θ = 11.41 revolutions

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