Question

A dentist's drill starts from rest. After 3.00 s of constant angular acceleration it turns at a rate of 2.60 x 104 rev/min (a) Find the drill's angular acceleration. rad/s (b) Determine the angle (in radians) through which the drill rotates during this period rad

Answer 1

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 $\frac{rev}{min}\frac{6.28rad}{1 rev}\frac{1 min}{60 sec}$ = 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