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# 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

- oldschoolLv 72 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)

- AshLv 72 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