Answer:
(a)
907.11 rad/s²
(b)
4082 rad
Explanation:
(a)
t = time taken = 3.00 s
w₀ = initial angular velocity of the drill = 0 rad/s
w = final angular velocity of the drill = 26000 rev/min = 26000 [tex]\frac{rev}{min}\frac{6.28rad}{1 rev}\frac{1 min}{60 sec}[/tex] = 2721.33 rad/s
α = angular acceleration of the drill
t = time interval = 3.00 s
using the equation
w = w₀ + α t
2721.33 = 0 + α (3)
α = 907.11 rad/s²
(b)
θ = angle
using the equation
θ = w₀ t + (0.5) α t²
θ = (0) (3.00) + (0.5) (907.11) (3.00)²
θ = 4082 rad
