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Anonymous
Anonymous asked in Science & MathematicsEngineering · 2 months ago

A dentist's drill starts from rest. After 2.90 s of constant angular acceleration it turns at a rate of 2.30 ✕ 104 rev/min.?

(a) Find the drill's angular acceleration.

 ___rad/s2

(b) Determine the angle (in radians) through which the drill rotates during this period.

 ___rad

3 Answers

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  • 2 months ago

    2.3e4 rev/min *2π rad/rev * 1min/60s = ω = 2409 rad/s

    α = dω/dt  ≈  Δω/Δt = 2409/2.9 = α = 831rad/s² (a)

    Θ = Θo + ωo*t + ½αt² = 0 + 0*t + ½ *831 *2.90² =  3492 rad or about 3490 rad (b)

  • Ash
    Lv 7
    2 months ago

    (a)

    ω₀ = 0

    ω = rpm * (1 min / 60 s) * 2𝜋 = (2.30 x 10⁴ rpm)*(1 min / 60 s)*2𝜋 = 2.41 x 10³ rad/s

    ω = ω₀ + αt

    α = (ω - ω₀)/t

    α = [(2.41 x 10³ rad/s) - 0]/(2.90 s)

    α = 831 rad/s²

    (b) ω² = ω₀² + 2αθ

    θ = (ω² - ω₀²)/(2α)

    θ = [(2.41 x 10³ rad/s)² - 0²]/(2*831 rad/s²)

    θ = 3.50 x 10³ rad

  • 2 months ago

    2.30 ✕ 104  ?? what does this mean ?

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