A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What is the linea

Question

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What is the linear speed (in m/s) of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2

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Khánh Gia 5 years 2021-07-15T16:03:54+00:00 1 Answers 16 views 0

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  1. thuthao
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    2021-07-15T16:05:49+00:00
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    Answer:

    The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

    Explanation:

    We are given that

    Angular acceleration, \alpha=3.3 rad/s^2

    Diameter of the wheel, d=21 cm

    Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

    Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

    1m=100 cm

    Magnitude of total linear acceleration, a=1.7 m/s^2

    We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

    Tangential acceleration,a_t=\alpha r

    a_t=3.3\times \frac{21\times 10^{-2}}{2}

    a_t=34.65\times 10^{-2}m/s^2

    Radial acceleration,a_r=\frac{v^2}{r}

    We know that

    a=\sqrt{a^2_t+a^2_r}

    Using the formula

    1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

    Squaring on both sides

    we get

    2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

    \frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

    v^4=r^2\times 2.7699

    v^4=(10.5\times 10^{-2})^2\times 2.7699

    v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

    v=0.418 m/s

    Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

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