At a certain instant, a rotating turbine wheel of radius RR has angular speed ωω (measured in rad/srad/s). What must be the magnitude αα of

Question

At a certain instant, a rotating turbine wheel of radius RR has angular speed ωω (measured in rad/srad/s). What must be the magnitude αα of its angular acceleration (measured in rad/s2rad/s2) at this instant if the acceleration vector a⃗ a→ of a point on the rim of the wheel makes an angle of exactly 30∘∘ with the velocity vector v⃗ v→ of that point? Express your answer in terms of some or all of the variables RRR and ωωomega.

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Thu Nguyệt 1 week 2021-07-22T20:32:02+00:00 1 Answers 1 views 0

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    2021-07-22T20:33:32+00:00

    Answer:

    Explanation:

    The magnitude of the acceleration makes an angle of 30° with the tangential velocity.

    Resolving the acceleration to tangential and radial acceleration

    at = aCos30 = √3a/2

    ar = aSin30 = ½a

    a = 2•ar

    Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:

    at = Rα

    Then, α = at/R

    since at = √3a/2

    Then, α = √3 at/2R, equation 1

    The radial acceleration is given as

    ar = ω²R

    Note that, at² + ar² = a²

    at = √(a²-ar²)

    Back to equation 1

    α = √3 at/2R

    α = √3√(a²-ar²)/2R

    α = √3√(a²-(w²R)²)/2R

    α = √3(a²-w⁴R²) / 2R

    Also, a = 2•ar = 2w²R

    Then,

    α = √3((2w²R)²-w⁴R²) / 2R

    α = √3(4w⁴R²-w⁴R²) / 2R

    α = √3(3w⁴R²) / 2R

    α = √9w⁴R² / 2R

    α = 3w²R / 2R

    α = 3w²/2

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