A car of mass 410 kg travels around a flat, circular race track of radius 83.4 m. The coefficient of static friction between the wheels and

Question

A car of mass 410 kg travels around a flat, circular race track of radius 83.4 m. The coefficient of static friction between the wheels and the track is 0.286. The acceleration of gravity is 9.8 m/s 2 . What is the maximum speed v that the car can go without flying off the track

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Thiên Thanh 3 years 2021-08-26T08:38:42+00:00 1 Answers 114 views 0

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  1. kimcuc
    0
    2021-08-26T08:39:50+00:00
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    Answer:

    The maximum speed v that the car can go without flying off the track = 15.29 m/s

    Explanation:

    let us first lay out the information clearly:

    mass of car (m) = 410 kg

    radius of race track (r) = 83.4 m

    coefficient of friction (μ) = 0.286

    acceleration due to gravity (g) = 9.8 m/s²

    maximum speed = v m/s

    For a body in a constant circular motion, the centripetal for (F) acting on the body is given by:

    F = mass × ω

    where:

    F = maximum centripetal force = mass × μ × g

    ω = angular acceleration = \frac{(velocity)^2}{radius}

    ∴ F = mass × ω

    m × μ × g = m × \frac{v^{2} }{r}

    410 × 0.286 × 9.8 = 410 × \frac{v^{2} }{83.4}

    since 410 is on both sides, they will cancel out:

    0.286 × 9.8 = \frac{v^{2} }{83.4}

    2.8028 = \frac{v^{2} }{83.4}

    now, we cross-multiply the equation

    2.8028 × 83.4 = v^{2}

    v^{2} = 233.754

    ∴ v = √(233.754)

    v = 15.29 m/s

    Therefore, the maximum speed v that the car can go without flying off the track = 15.29 m/s

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