Ends of a chord ( T and S ) slide along the circumference, such that chord TS always remains horizontal ....... ?

Ends of a chord ( T and S ) slide along the circumference, such that chord TS always remains horizontal. At an instant, TS is descending vertically at a uniform rate of (0.1 r) per second, find rate of change of Area ratio of the two segments when the chord is at a height of (r/2) above the center.

Update:

Descending rate of the chord is uniform at any instance.

Update 2:

Pope: Hi, your approach is always appreciable. I too had calculated this rate. My answer was 0.082699/s,   which exactly the same that you have calculated. Plz allow me to keep this question open for some time. Regards.

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  • Pope
    Lv 7
    2 months ago
    Favorite Answer

    By the ratio of the two sectors, I suppose you mean minor and major sectors TS.

    Referring to your sketch, when the chord is r/2 above center, θ = π/3. The central angle of the minor sector is 2π/3.

    The height of the chord above center is rcos(θ).

    d/dt rcos(θ) = -0.1r

    -rsin(θ) dθ/dt = -0.1r

    dθ/dt = 1/[10sin(θ)]

    The ratio of the sector areas is equal to the ratio of their central angles. Let w be that ratio.

    w = area(minor TS) : area(major TS)

    w = 2θ / (2π - 2θ)

    w = θ / (π - θ)

    When θ = π/3,

    w = 1/2

    dθ/dt = 1/[10sin(π/3)] = √(3)/15

    w = θ / (π - θ)

    πw - θw = θ

    Differentiate with respect to t.

    πw' - θw' - θ'w = θ'

    w' = (θ' + θ'w) / (π - θ)

    w' = θ'(1 + w) / (π - θ)

    w' = [√(3)/15](1 + 1/2) / (π - π/3)

    w' = 3√(3) / (20π)

    The ratio of areas changes at rate 3√(3)/(20π) per second.

    Incidentally, your update states that the rate of vertical displacement of the chord is uniform. That really does not matter. For this question the instantaneous rate is all that matters.

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