A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man’s shoes and the merry-

Question

A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man’s shoes and the merry-go-round is µs = 0.5, how far from the axis of rotation can he stand without sliding?

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Acacia 5 years 2021-08-18T05:40:09+00:00 1 Answers 21 views 0

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  1. thugiang
    0
    2021-08-18T05:42:04+00:00
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    Answer:

    0.8 m

    Explanation:

    Draw a free body diagram.  There are three forces:

    Weight force mg pulling down,

    Normal force N pushing up,

    and friction force Nμ pushing towards the center.

    Sum of forces in the y direction:

    ∑F = ma

    N − mg = 0

    N = mg

    Sum of forces in the centripetal direction:

    ∑F = ma

    Nμ = m v²/r

    Substitute and simplify:

    mgμ = m v²/r

    gμ = v²/r

    Write v in terms of ω and solve for r:

    gμ = ω²r

    r = gμ/ω²

    Plug in values:

    r = (10 m/s²) (0.5) / (2.5 rad/s)²

    r = 0.8 m

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