Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father exerts a for

Question

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father exerts a force on the merry-go-round perpendicular to its radius to achieve an angular acceleration of 4.44 rad/s^2.

Required:
a. How long (in s) does it take the father to give the merry-go-round an angular velocity of 1.53 rad/s? (Assume the merry-go-round is initially at rest.)
b. How many revolutions must he go through to generate this velocity?
c. If he exerts a slowing force of 270 N at a radius of 1.20 m, how long (in s) would it take him to stop them?

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Mộc Miên 4 years 2021-08-02T07:23:08+00:00 1 Answers 36 views 0

Answers ( )

  1. King
    0
    2021-08-02T07:24:54+00:00
    Reply

    Answer:

    Explanation:

    Given that:

    the initial angular velocity \omega_o = 0

    angular acceleration \alpha = 4.44 rad/s²

    Using the formula:

    \omega = \omega_o+ \alpha t

    Making t the subject of the formula:

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

    where;

    \omega = 1.53 \ rad/s^2

    ∴

    t= \dfrac{1.53-0}{4.44 }

    t = 0.345 s

    b)

    Using the formula:

    \omega ^2 = \omega _o^2 + 2 \alpha \theta

    here;

    \theta = angular displacement

    ∴

    \theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

    \theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

    \theta =0.264 \ rad

    Recall that:

    2π rad = 1 revolution

    Then;

    0.264 rad = (x) revolution

    x = \dfrac{0.264 \times 1}{2 \pi}

    x = 0.042 revolutions

    c)

    Here; force = 270 N

    radius = 1.20 m

    The torque = F * r

    \tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm

    However;

    From the moment of inertia;

    Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}

    given that;

    I = 84.4 kg.m²

    \alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2

    For re-tardation; \alpha=-3.84 \ rad/s^2

    Using the equation

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

    t= \dfrac{0-1.53}{ -3.84 }

    t= \dfrac{1.53}{ 3.84 }

    t = 0.398s

    The required time it takes= 0.398s

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