A rotating wheel with diameter 0.800 m is speeding up with constant angular acceleration. The speed of a point on the rim of the wheel incre

Question

A rotating wheel with diameter 0.800 m is speeding up with constant angular acceleration. The speed of a point on the rim of the wheel increases from 3.00 m/s to 6.00 m/s while the wheel turns through 4.00 revolutions.

Required:
What is the angular acceleration of the wheel?

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King 4 years 2021-08-21T14:58:05+00:00 1 Answers 29 views 0

Answers ( )

  1. Amity
    0
    2021-08-21T15:00:00+00:00
    Reply

    Answer:

    The angular acceleration of the wheel is 3.357rad/s²

    Explanation:

    Given;

    Diameter of the wheel, d = 0.8 m

    radius of the wheel, r = 0.4 m

    initial velocity of the wheel, u = 3 m/s

    final velocity of the wheel, v = 6 m/s

    number of revolutions of the wheel, N = 4 rev.

    Convert the linear velocity in m/s  to angular velocity in rad/s;

    v = ωr

    ω = v / r

    initial angular velocity, \omega_i = \frac{3}{0.4} = 7.5 \ rad/s

    final angular velocity, \omega_f= \frac{6}{0.4} = 15\ rad/s

    The angular acceleration of the wheel is calculated as;

    \omega _f^2 = \omega_i^2 + 2 \alpha \theta

    where;

    α is the angular acceleration (rad / s²)

    θ is the angular rotation (rad)

    θ = Number of revolutions x 2π rad/rev

      =  4 rev. x 2π rad/rev

      = 25.136 rad.

    \omega _f^2 = \omega_i^2 + 2 \alpha \theta\\\\2 \alpha \theta = \omega _f^2 - \omega_i^2\\\\\alpha = \frac{\omega _f^2 - \omega_i^2}{2 \theta} \\\\\alpha = \frac{15^2 -7.5^2}{2*25.136} \\\\\alpha = 3.357 \ rad/s^2

    α = 3.357 rad/s²

    Therefore, the angular acceleration of the wheel is 3.357rad/s²

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