A wheel rotating with a constant angular acceleration turns through 23 revolutions during a 3 s time interval. Its angular velocity at the e

Question

A wheel rotating with a constant angular acceleration turns through 23 revolutions during a 3 s time interval. Its angular velocity at the end of this interval is 16 rad/s. What is the angular acceleration of the wheel

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Mộc Miên 5 years 2021-07-31T05:27:29+00:00 2 Answers 22 views 0

Answers ( )

  1. thuthuy
    0
    2021-07-31T05:28:49+00:00
    Reply

    Answer

    Given,

    Revolution of wheel = 23 rev.

                                      = 23 x 2π = 46 π

    Time = 3 s

    final angular velocity = 16 rad/s

    angular acceleration of wheel = ?

    Now, Calculating the initial angular speed of the wheel

    Angular displacement = \dfrac{1}{2}(initial velocity + final velocity) x time.

    46 π = \dfrac{\omega_o+16}{2}\times 3

    \omega_0 = 80.29\ rad/s

    now, angular acceleration

    \alpha = \dfrac{\omega-\omega_0}{t}

    \alpha = \dfrac{16-80.29}{3}

    \alpha = -21.43\ rad/s^2

    Hence, the angular acceleration of wheel is negative means wheel is decelerating.

  2. kimcuc
    0
    2021-07-31T05:28:58+00:00
    Reply

    Answer:

    – 21.4 rad/s²

    Explanation:

    number of turn, n = 23

    angle turned, θ = n x 2π = 23 x 2 x π = 46 π rad

    time, t = 3 s

    final angular velocity, ω = 16 rad/s

    Let the initial angular velocity is ωo.

    Angular displacement = Average angular velocity x time

    \theta = \frac{(\omega + \omega_{0})}{2}\times t

    46π  x 2 = (16 + ωo) x 3

    ωo = 80.3 rad/s

    Let α is the angular acceleration, by use of first equation of motion for rotational motion

    \omega = \omega_{0}+\alpha\times t

    16 = 80.3 + α x 3

    α = – 21.4 rad/s²

    The angular acceleration of the wheel is – 21.4 rad/s².

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