A 2.3kg bicycle wheel has a diameter of 50cm. What torque must you apply to take the wheel from 0rpm to 120rpm in 5.5s?

Question

A 2.3kg bicycle wheel has a diameter of 50cm. What torque must you apply to take the wheel from 0rpm to 120rpm in 5.5s?

in progress 0
Delwyn 1 year 2021-08-18T18:11:41+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-08-18T18:13:25+00:00

    Answer:

    τ = 0.26 N.m

    Explanation:

    First we find the moment of inertia of the wheel, by using the following formula:

    I= mr²

    where,

    I = Moment of Inertia = ?

    m = mass of wheel = 2.3 kg

    r = radius of wheel = 50 cm/2 = 25 cm = 0.25 m

    Therefore,

    I = (2.3 kg)(0.25 m)²

    I = 0.115 kg.m²

    Now, we find the angular acceleration of the wheel:

    α = (ωf – ωi)/t

    where,

    α = angular acceleration = ?

    ωf = final angular velocity = (120 rpm)(2π rad/1 rev)(1 m/60 s) = 12.56 rad/s

    ωi = Initial Angular Velocity = 0 rad/s

    t = time = 5.5 s

    Therefore,

    α = (12.56 rad/s – 0 rad/s)/(5.5 s)

    α = 2.28 rad/s²

    Now, the torque is given as:

    Torque = τ = Iα

    τ = (0.115 kg.m²)(2.28 rad/s²)

    τ = 0.26 N.m

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )